Ray Optics and Optical Instruments

CLASS XII

PHYSICS

By:DR. VIKRAM SINGH TANUSHREE SINGH

YEAR OF PUBLICATION-2010 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, transmitted in any form or by any meansElectronic, Mechanical, Photocopying, Recording or otherwise, without prior permission of the Authors and Publisher

TM

SAVANT INSTITUTE

Physics

Ray Optics & Optical Instruments

1

7 RAY OPTICS AND OPTICAL INSTRUMENTS §

Galileo Galilei (1564 – 1642), upon hearing rumors of the new device, experimented with lenses until he had built the first practical telescope in 1610. _____________________ Slide 4 ______________________

Moon from Apollo 16

_____________________ Slide 1 ______________________ Thin Lenses: History § §

§

Thin lens were first used for practical purposes by a Dutch merchant, Anton van Leeuwenhoek (1632 – 1723). He used very small pieces of glass (it is easier to have a flawless small piece of glass than a flawless large one) and polished them so accurately that he could get magnifications of more than 200 without loss of detail. He was able to see blood capillaries, and tiny living animals (protozoa).

_____________________ Slide 5 ______________________ Stamp in the honour of Robert Snell

_____________________ Slide 2 ______________________ Microscopes § § §

Such strong magnifying lenses are microscopes (from Greek words meaning “to see the small”). A microscope, like the one Leeuwenhoek used, made with one lens are called “simple” microscopes. If two lenses are used it is called a “compound” microscope. _____________________ Slide 3 ______________________ _____________________ Slide 6 ______________________

Telescopes § §

The word telescope comes from the Greek “to see the distant. The telescope is supposed to have been invented by an apprentice-boy in the shop of the Dutch spectacle maker Hans Lipershey (ca. (1570 – 1619) in about 1608.

Introduction: §

§

Light is a non-mechanical (requires no medium for propagation) form of energy due to which we have sensation of vision. The branch of study of light is called optics.

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94 §

Ray Optics & Optical Instruments Broadly optics is divided into three groups. (1) Geometrical optics (2) Wave optics (3) Quantum optics

_____________________ Slide 7 ______________________ Geometrical optics §

In this, light is considered as a ray on which its transfer takes place. _____________________ Slide 8 ______________________

§ § §

Physics Slide 11

What Can Happen to Light? When it strikes a surface of an object light can be: 1.

Reflected

2.

Transmitted

3.

Absorbed (and transformed to thermal energy)

_____________________ Slide 12 _____________________ Reflected

Geometrical optics states that for each and every object there is an image. Geometrical optics works on following assumptions. Rectilinear propogration of light i.e. light ray travels in in straight line.

1. Lawsof reflection. 2. Lawsof refraction

3.

Physical independence of light rays i.e. two light rays are totally independent from each other.

_____________________ Slide 13 _____________________ Transmitted

_____________________ Slide 9 ______________________ The Ray (Particle) Model of Light § § § § § §

Evidence suggests that light travels in straight lines under a wide variety of circumstances. We infer the positions of objects by assuming h t at light moves from the objects to our eyes in straight lines. This is the ray model of light. Newton used the ray model. Ray model explains reflection, refraction, and the formation of images by mirrors and lenses. This subject is often referred to as geometrical optics.

_____________________ Slide 14 _____________________ Absorbed

_____________________ Slide 10 _____________________ Ray Model

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Physics

Ray Optics & Optical Instruments Slide 15 Reflection and Scattering

Slide 19 Reflection of Light

_____________________ Slide 16 _____________________ Movie screen scatters light in all directions

95

_____________________ Slide 20 _____________________ Object §

Any source of light is called an object and is of two types.

Real object §

When light rays diverse from a light source, object is real.

_____________________ Slide 21 _____________________ _____________________ Slide 17 _____________________ Interactions of Light with Matter

Virtual object §

When light rays converge towards the light source, object is virtual.

§

To find the position of object we should find out the point of intersection of incident rays. _____________________ Slide 22 _____________________

Image Interactions between light and matter determine the appearance of everything around us: objects reflect some wavelengths, absorb others and emit others. _____________________ Slide 18 _____________________ The Speed of Light and Index of Refraction §

The accepted value today for the speed of light in a vacuum is c = 2.99792458 x 108 m/s. We usually round off to c = 3.0

§

To find the position of Image we should find out the point of intersection of reflected or refracted ray. There are of two types of images.

Real Image §

If the reflected or refracted ray converges towards the point of intersection the image is real.

x 108 m/s.

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Ray Optics & Optical Instruments Slide 23

Physics Slide 28

Virtual Image §

If the reflected or refracted rays diverges from a point then the point of intersection is virtual Image.

_____________________ Slide 24 _____________________ Mirror §

A smooth and shiny surface on which reflection can take place is called a mirror, having two types. 1. Plane mirror. 2. Spherical mirror. _____________________ Slide 25 _____________________

Reflection of light § § § §

_____________________ Slide 29 _____________________ Image Formation—Plane Mirror

The returing back of light in the same medium from which it has come after striking a surface is called reflection of light. Reflection may be Diffused reflection – here random reflections take place through irregular surfaces. Specular reflection - here regular reflections take place through plane surfaces. _____________________ Slide 26 _____________________

Reflection

_____________________ Slide 30 _____________________ Image Formation –Plane Mirror

_____________________ Slide 27 _____________________ Diffuse reflection

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97

Slide 31

_____________________ Slide 35 _____________________ Image of a Point in a Plane Mirror OA = OA′

_____________________ Slide 32 _____________________

_____________________ Slide 36 _____________________

Conceptual Example How tall must a full-length mirror be? A women 1.60 m tall stands in front of a vertical plane mirror. What is the minimum height of the mirror, and how high must it lower edge be above the floor, if she is to see her whole body? (Assume her eyes are 10 cm below the top of her head.)

Image of an extended object in a plane The image will be an inverted one.

_____________________ Slide 33 _____________________ _____________________ Slide 37 _____________________ Properties of the image formed by a plane mirror § § §

The image is as far behind the mirror as the object is in front of the mirror. Object distance = image distance. The image is unmagnified, virtual and erect. Height of object = height of Image. The image has front-back reversal. i.e. It shows lateral inversion i.e. left potion of object appears as right portion of Image. _____________________ Slide 38 _____________________

_____________________ Slide 34 _____________________

§

A plane mirror deviates light through an angle δ = 180° − 2i

§

Where i is the angle of incidence. The deviation is maximum for normal incidence. δmax = 180°. Glancing angle φ = angle between mirror and reflected rays is called glancing angle φ = 90 – i = 90 – r.

Reflection of light though plane mirror Laws of Reflection (i)

The incident ray, the reflected ray and the normal to the reflecting surface at the point of incidence all lie in the same

§

plane. (ii) The angle of reflection r is equal to the angle of incidence i, that is i = r.

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Ray Optics & Optical Instruments Slide 39

Rotation of Mirror §

If the direction of the incident ray is kept constant and the mirror is rotated through an angle θ about an axis in the plane of mirror, then the reflected ray rotates through an angle 2θ.

Physics Slide 43

Solved Example If,

θ = 90°, n = 3,

θ = 72°, n =

360° =5 72

If the object is placed symmetrically midway between the mirror, then 360° n= − 1 = 4. 72

_____________________ Slide 44 _____________________ Illustration _____________________ Slide 40 _____________________ §

If an object moves toward or away from) a plane mirror at speed v, the image will also approach (or recede) at the same speed v, and the relative velocity of image with respect to object will be 2v, as shown in figure a.

Two plane mirrors are inclined at an angle of 60°. An object is placed between the mirrors. What is the total number of images formed by two mirrors? _____________________ Slide 45 _____________________ Illustration A man H m tall wishes to see his full-length image in a plane mirror hanging vertically on a wall. Find the length of the shortest mirror which he can see his entire image. If his eyes are H m above the ground, find the position of the mirror. _____________________ Slide 46 _____________________

_____________________ Slide 41 _____________________ §

Spherical Mirrors

If the mirror is moved toward (or away from) the object with speed v, the image will also move towards or away from) the object with a speed 2v. _____________________ Slide 42 _____________________

Number of image by inclined mirrors § §

§ §

When θ, the angle between the two mirrors, is an exact sub multiple of 180° The total number of images formed due to successive 360°  360°  reflection is equal to either or  θ − 1 . θ   360° Accordingly as is odd or even, respectively. θ 360° In the first case ( is odd) when the object is placed θ exactly midway between the two mirrors, the two images  360°  coincide and the total number of images is  − 1 .  θ 

_____________________ Slide 47 _____________________

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Physics

Ray Optics & Optical Instruments Slide 48

99 Slide 51

(iii) Pole (P) It is the geometrical centre of the spherical reflecting surface. (iv) Principal Axis It is the straight line joining the centre of curvature to the pole.

_____________________ Slide 52 _____________________

_____________________ Slide 49 _____________________ Rays parallel to the principal axis of the mirror come to a focus at F, called the focal point, as long as the mirror is small in diameter, d, as compared to its radius of curvature, r. In that case q will be small and the rays will cross each other at very nearly the same point.

(v) Focus (F) When a narrow beam of rays of light, parallel to the principal axis and close to it, is incident on the surface of a mirror, the reflected beam is found to converage or appear to diverge from a point on the principal axis. This point is called the focus. (vi) Focal length (f) It is the distance between the pole and the principal focus. For spherical mirrors f = R/2 _____________________ Slide 53 _____________________ Reflection of light from spherical (curved) sign convention § §

_____________________ Slide 50 _____________________

_____________________ Slide 54 _____________________

Reflection from Curved Surfaces Important Terms

Pole is taken to be the origin and the principal axis as the xaxis. The quantities u, v, R and f is positive if the corresponding point lies on the positive side of the origin (in the direction of incident light) and negative if it is on the negative side. (opposite to the direction of incident light).

§

(i)

Centre of Curvature (C) It is the centre of sphere of which the mirror is a part. (ii) Radius of Curvature (R) It is the radius of the spherical reflecting surface.

The distances measured in the upward direction, perpendicular to the principal axis of the mirror are taken as positive and those measured in downward direction is taken as negative. _____________________ Slide 55 _____________________

Rules for ray diagrams §

The position, size and nature of an image formed by mirrors are conventionally expressed as ray diagrams. We can locate the image of any extended object graphically by drawing any two of the following four special rays: (a) A ray, initially parallel to the principal axis is reflected through the focus of the mirror.

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Ray Optics & Optical Instruments Slide 56

§

(b) A ray, initially passing through the focus is reflected parallel to the principal axis. (c) A ray passing through the centre of curvature is reflected back along itself. (d) A ray incident at the pole is reflected symmetrically.

§

Physics

AP AP AP , β= and γ ≈ PO PC PI AP AP AP Hence , 2 = + PC PO PI 1 1 2 or, + = PO PI PC α=

Fig (i)

..... ( iv )

_____________________ Slide 61 _____________________

_____________________ Slide 57 _____________________ §

The pole P is taken as the origin and the principal axis as

In terms of Cartesian Sign convention mirror formula may be expressed as: 1 1 1 + = v u f

§

the X – axis. As, the distances PO, PI and PC are positive, PO = –u, PI =

Where,

Note:

Mirror Formula

u = object distance from the pole v = image distance from the pole,

– V and PC = – R. Putting in (iv) 1 1 2 1 1 2 1 + = = + = = –u –v –R u v R f

§

f = focal length of the mirror.

While using the mirror formula always write only known quantities with sign and unknown quantities without sign.

_____________________ Slide 58 _____________________

_____________________ Slide 62 _____________________

Relation between u, v and R for spherical mirrors Magnification §

Consider the situation shown in figure. A point object is placed at the point O of the principal axis of a concave mirror.

§

The line CA is the normal at A.

§

Magnification – It is defined as the ratio of dimension of

§

Image to the dimension of object (m). It is a pure number. _____________________ Slide 63 _____________________

Lateral magnification or transverse magnification

_____________________ Slide 59 _____________________ §

Thus, by the laws of reflection. ∠OCA = ∠CAI. Let α, β, γ

It is given by, m =

§

Where h i = height of image,

§

If magnification m is positive, the image is erect w.r.t. respect to the object;

§

If m is negative image is inverted with respect to the object.

§ §

For a real image by spherical mirror, m is negative. For a virtual image by spherical mirror, m is positive.

and θ denote the angles AOP, ACP, AIP and OAC §

Application of mirror formula

opposite interior angles, we have,

§

From triangle

OAC β = α + θ

….(i)

And from triangle

OAI γ = α + 2θ

….(ii)

Eliminating θ are from (i) and (ii) 2β = α + γ

§

…..(iii)

h0 = height of object.

_____________________ Slide 64 _____________________

respectively. As the exterior angle in a triangle equals the sum of the

_____________________ Slide 60 _____________________ §

hi − v = h0 u

§

The mirror formulae may be used to find the position, nature and size of the image formed by a spherical mirror.

Limitation §

The limitation of the formulae is that it is applicable only for paraxial rays (the rays which make very small angle with the principal axis).

If the point A is close to P, the angles α, β and γ are small and we can write.

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Ray Optics & Optical Instruments

101 Slide 69

Slide 65 Relation between f and R – §

From mirror formula we have derived

1 1 2 + = , now if the u v R

object is at infinity the image is formed at focus. Hence, when u = ∞ v = f, Now from mirror formula 1 1 2 1 2 + = or = or, f = R/2 i.e. focal length of a mirror is f ∞ R f R half of its radius of curvature.

§

_____________________ Slide 66 _____________________

_____________________ Slide 70 _____________________

Object Not at Infinity

(a) Ray 1 goes out from O′ parallel to the axis and reflects through F. _____________________ Slide 67 _____________________

_____________________ Slide 71 _____________________ Convex Mirrors

(b) Ray 2 goes through F and then reflects back Parallel to the axis. _____________________ Slide 68 _____________________

_____________________ Slide 72 _____________________

(c)

Ray 3 heads out perpendicular to mirror and the reflects back on itself and goes through C center of curvature SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

102

Ray Optics & Optical Instruments

Physics Slide 78

Slide 73 Illustration

Introduction:

An object of length 3 cm is placed at a distance of (5/4)f from a concave mirror where f is the magnitude of the focal length of the mirror. The length of the object is parallel to the principal axis. Find the length of the image, is the image erect or inverted?

Refraction through plane surface refraction of light § §

Refraction is a phenomena of light due to which it bends while travelling from one medium to another. When light goes from one medium to another medium. ü Its velocity changes ü Its wavelength changes ü Its path may or may not change. ü Its frequency remains unchanged. _____________________ Slide 79 _____________________

§ _____________________ Slide 74 _____________________ Newton’s formula If the object and image distances are measured from the focus instead of the pole of the mirror. Then, the mirror formula reduces to a simple form called the Newton’s formula. x0 xi = f 2 Where x0 is the object distance from the focus. xi is the image distance from the focus.

§

During, the propagation of light from one medium to another medium changes its velocity and wavelength and the path also changes (except ∠i = 0). When light goes from an optically rarer to an optically dense medium, it bends towards normal provided ∠i ≠ 0 as shown in the fig.

_____________________ Slide 75 _____________________ Illustration A short linear object is placed at a distance ‘u’ along the axis of a spherical mirror of focal length f. (i) Obtain an expression for the longitudinal magnification. (ii) Also, obtain an expression for the ratio of the velocity of image (v) to the velocity of object (u). _____________________ Slide 76 _____________________ Illustration

_____________________ Slide 80 _____________________ When light goes from an optically dense to an optically rarer medium it bends away from normal provided. ∠i ≠ 0, Fig.

A small strip of plane mirror A is set with its plane normal to the principal axis of a convex mirror B and placed 15 cm in front of B which it partly covers. An object is placed 30 cm from A and the two virtual images formed by reflection in A and B coincide without parallax. Find the radius of curvature of B. _____________________ Slide 77 _____________________ Illustration A concave mirror of focal length 15 cm) and a convex mirror (focal length 10 cm) are placed co-axially 70 cm apart facing each other. A 2 cm tall object is placed perpendicular to the common axis 20 cm from the concave mirror. Find the position, size and nature of the final image formed by two reflections, first at concave mirror and then at convex mirror. SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

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Ray Optics & Optical Instruments

103 Slide 84

Slide 81

_____________________ Slide 85 _____________________ Refractive index

_____________________ Slide 82 _____________________

§

The degree of bending of light depends upon the medium quality called refraction Index (µ), a pure number.

§

Refractive index (µ) is of two types 1. 2.

Relative Refractive Index (µ) Absolute Refractive Index

_____________________ Slide 86 _____________________ §

The absolute refractive index (µ) of a medium is defined as the ratio of the speed of light in the vacuum (c) to the speed of light in the medium (Vm). c µ= Vm _____________________ Slide 87 _____________________

Relative refractive index _____________________ Slide 83 _____________________

§

The relative refractive index of two media is equal to the ratio of their absolute refractive indices. µ c v2 v1 µ21 = 2 = = µ1 c v1 v 2

§ §

This is read as refractive index of 2nd medium with respect 1st medium and written as 1µ2. 3 The refractive indices of glass and water are ( = 1.5) and 2 4 ( = 1.33 ) . 3 _____________________ Slide 88 _____________________

Note : §

Absolute Refractive index µ is the refractive index of a medium with respect to air or vacuum, i.e. c λ air µ= = where, λ denotes wavelength of light. Vm λ medium

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104

Ray Optics & Optical Instruments Slide 92

Slide 89 Refraction of Light (Snell's Law) §

Physics

§

As light travels from one medium to another, its frequency does not change. §

As the incidence angle (θ1) is increased until a particular angle (θc), as shown in figure, the angle of refraction will be 90º and the refracted ray would skim the surface of the glass. The incidence angle at which this occurs is called the critical angle, denoted as

_____________________ Slide 90 _____________________ ü

Both the wave speed and the wavelength do change.

ü

The ratio of the indices of refraction of the two media can be expressed as various ratios sin θ1 n2 v 1 λ 1 1n2 = = = = sin θ2 n 1 v 2 λ 2

_____________________ Slide 93 _____________________ Total internal reflection

_____________________ Slide 91 _____________________ Refraction of Light Critical Angle §

When light attempts to move from a medium having a given

“Total internal reflection is an effect occurs only when light

index of refraction to one having a lower index of refraction.

attempts to move from a medium of given index of refraction to a medium of lower index of refraction and the incidence angle is equal or greater than the critical angle of the higher refractive index medium .” _____________________ Slide 94 _____________________ Total internal reflection § §

We can use Snell’s law of refraction to find the critical angle. When θ1 = θ c , θ 2 = 90º and Snell’s law gives n1 sin θc = n2 sin90O = n2 n n sin θc = 2 ⇒ θc = sin−1 2 n1 n1 For n1 > n2

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Physics

Ray Optics & Optical Instruments Slide 95

105 Slide 98

§

Example Calculate the critical angle of a diamond (n = 2.42) 1 sin θc = nD 1 sin θc = 2.42 θ c = 24.4º

§

Light, signals or other forms of communication can travel a long distance without losing much intensity. Applications include medical use of fiber optic cables for diagnosis and correction of medical problems Telecommunications _____________________ Slide 99 _____________________

Application of total internal reflection §

Total internal reflection of light by prisms in binoculars .

_____________________ Slide 96 _____________________

_____________________ Slide 100 ____________________ Application of total internal reflection

_____________________ Slide 97 _____________________ Fiber Optics Application of total internal reflection

_____________________ Slide 101 ____________________

§

Such a “light pipe” is flexible if thin fibers are used rather than thick rods. If a bundle of parallel fibers is used to construct an optical transmission line, images can be transferred from one point to another.

Diamonds achieve their brilliance from a combination of dispersion and total internal reflection. Because diamonds have a very high index of refraction of about 2.4, the critical angle for total internal reflections only 25 degree. Incident light therefore strikes many of the internal surfaces before it strikes one at less than 25 degree and emerges. After many such reflections, the light has traveled far enough that the colors have become sufficiently separated to be seen individually and brilliantly by the eye after leaving the crystal.

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106

Ray Optics & Optical Instruments

Physics

µ1 sin θ1 = µ2 sin θ2

Slide 102 Some Indices of Refraction

or,

sin θ 1 µ 2 = >1 sin θ 2 µ1

Thus, if µ2 > µ1 the θ2 < θ1. i.e light ray bends towards the

§

normal. _____________________ Slide 106 ____________________ §

When a light ray passes from denser to rarer medium it

§

bends away from the normal as shown in the figure. From Snell’s Law, we know that sin θ 1 µ 2 = > 1P sin θ 2 µ1

_____________________ Slide 103 ____________________ Snell’s law §

plane. §

_____________________ Slide 107 ____________________

When light propagates through a series of layers of different medium as shown in the figure, then according to Snell’s law

Refraction: Principle of Least

we may write,

Time,

µ1 sin θ1 = µ2 sin θ2 = µ3 sin θ3 = … = constant _____________________ Slide 104 ____________________ §

Thus, if µ2 < µ 1, θ2 > θ .i.e. refracted ray bends away form the normal.

§

Incident ray, Refracted ray and Normal lie in the same

In general, µ sin θ = constant. i.e. for given any two media where light ray strikes the

t = t1 + t2

=

s1 s2 + v1 v 2

v=

c , n

t=

1 dt 2 2 1/2 [n(y + n2( y 2 2 + ( a - x )2 )1/2 ] =0 1 1 + x ) c dx

interface

t=

n1s1 n2s 2 + c c

µ1 × sin i = µ2 × sin r

Prove Snell’s law, n1sin?1= n2sin ?2. Demonstrate law of reflection. _____________________ Slide 108 ____________________ Relation between object and image distance §

An object O placed in medium 1 (refractive index µ1) is viewed from the medium 2 (refractive index µ2).

_____________________ Slide 105 ____________________ Conclusion from Snell’s law §

When light passes from rarer to denser medium, it bends towards the normal as shown in the figure.

§

Using Snell’s Law at point P

§

It is important to note that the object and image both are formed on the same side of the boundary.

§

The image distance y and the object distance x are related as µ  y=  2 x  µ1 

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Physics

Ray Optics & Optical Instruments

107 Slide 113

Slide 109 §

For an angle of incidence greater than θc, the light is totally reflected back into the medium of higher refractive index. This phenomenon is called total internal reflection.

_____________________ Slide 114 ____________________ Optical fibers: § Here,

x = Real depth/height y = Apparent depth/height

The optical fibres can transmit light beam from one end to the other due to the repeated total internal reflections even if the fiber is bent or twisted.

_____________________ Slide 110 ____________________ §

§

If µ2 > µ1, that is, when the object is observed from a denser medium, it appears to be farther away from the interface, i.e., y > x. or, Apparent height > Real height. If µ 2 < µ1, that is, when the object is observed from a rarer medium, it appears to be closer to the interface, i.e. y < x. i.e. Apparent depth < Real depth.

_____________________ Slide 115 ____________________

Note:

Mirage:

§

§

The above formula is only applicable for normal view or paraxial ray assumption. _____________________ Slide 111 ____________________

§

The optical illusion that water is present at some distance place is called inferior mirage. This generally occurs on very hot summer days . This is due to total internal reflection.

Total Internal Reflection

Looming:

§

§

§

The phenomenon of total internal reflection occurs when light travels from a medium of high refractive index to a medium of lower refractive index. At the critical angle (θc), the refracted ray just grazes the boundary between two media. _____________________ Slide 112 ____________________

§

Using Snell’s law, we get

§

µ  µ1 sin θc = µ2 sin 90° or , θ = sin −1  2   µ1 

§

Here, θ1 < θc < θ 2

§

The optical illusion of object floating in air is called superior mirage. It is also known as looming. This occurs in very cold regions due to total internal reflection. _____________________ Slide 116 ____________________

Total Internal Reflection; Fiber Optics

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108

Ray Optics & Optical Instruments Slide 117

Physics Slide 122

Refraction through glass slab §

_____________________ Slide 118 ____________________

§ §

The refracting surfaces of a glass slab are parallel to each other. When a light ray travels through a glass slab, it is refracted twice at the two parallel faces and finally emerges out parallel to its incident direction. The light ray undergoes zero deviation, δ = O. Angle of emergence = Angle of incidence Or, e = i _____________________ Slide 123 ____________________

§

The lateral displacement of the ray is the perpendicular distance between the incident and the emergent ray and is given by

 t  d=  sin (i − r )  cos r  _____________________ Slide 119 ____________________ Deviation ( δ ) §

The figure shows a light ray traveling from a denser to a rarer medium at an angle α lesser than the critical angle θc.

§

The deviation δ of the light ray is given by _____________________ Slide 124 ____________________

_____________________ Slide 120 ____________________ Conceptual Example View up from under water Describe what a person would see who looked up at the world from beneath the perfectly smooth surface of a lake or swimming pool. _____________________ Slide 121 ____________________ _____________________ Slide 125 ____________________ Apparent shift When a glass slab of thickness t and refractive index µ is placed in the path of a convergent beam as shown in the figure, then the point of convergence is shifted by SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

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Ray Optics & Optical Instruments

 1 s = t  1−   µ

_____________________ Slide 126 ____________________ §

When the same glass slab is placed in the path of a diverging beam, the point of divergence is  1 s = t  1−   µ

§

It is important to note that the shift (s) is always on the direction of light.

109 Slide 129

Prism §

Prism is a transparent medium with two refracting surfaces through which refraction takes place.

§

The two faces are not parallel but are inclined to each other.

§

Any geometrical figure with two refracting surfaces are not parallel in a prism.

§

The above figure is an example of prism.

§

In general, we take a triangular prism.

_____________________ Slide 130 ____________________ Visible Spectrum and Dispersion

_____________________ Slide 131 ____________________ n Dependence on Wavelength _____________________ Slide 127 ____________________ §

If there are n number of slabs with different refractive indices are placed between the observer and the object, then the total apparent shift is equal to the summation of the individual shifts. s = s1 + s 2 + ……… + sn  1  1  1 or, s = t 1  1-  + t 2 1-  +......t n  1-   µ1   µ2   µn  _____________________ Slide 128 ____________________

Illustration In the figure, determine the apparent shift in the position of the coin. Also, find the effective refractive index of the combination of the glass and water slab.

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110

Ray Optics & Optical Instruments Slide 132

Physics Slide 136

Newton’s Analysis of Light λ1 n2 = λ 2 n1

_____________________ Slide 137 ____________________ Basic Terms Angle of prism or reflecting angle (A) § _____________________ Slide 133 ____________________

§

The angle between the refracting faces is calledangle of prism. The adjoining figures shows various terms related to prism. _____________________ Slide 138 ____________________

Dispersion

Here,

i = angle of incidence r1 = angle of refraction through 1 st surface. r2 = angle of refraction through 2 nd surface. e = angle of emergence δ = angle of deviation

_____________________ Slide 134 ____________________

_____________________ Slide 139 ____________________ Angle of deviation (δ) _____________________ Slide 135 ____________________

§ §

It is the angle between the emergent and the incident ray. It other words , it is the angle through which incident ray turns in passing through a prism.

§

Now, the angle of deviation is given by δ = (i – r1) + (e + r 2) or,

δ = i + e – (r1 + r 2)

Also, ∠A = r1 + r 2 or, δ = ( i + e – A)

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Physics

Ray Optics & Optical Instruments Slide 140

sin(

Condition of minimum deviation §

§ §

At the minimum deviation the angle of incidence is equal to the angle of emergence i.e., i = e Also, r1 = r2 = r A = 2r The refracted ray inside the prism becomes parallel to the base of the prism. The deviation of the prism is given by or, Where δm is the minimum deviation produced. δ +A i= m 2 _____________________ Slide 141 ____________________

§

Using Snell’s law sin i µ= = sin r

A + δm A + δm ); 2 2

and sin

A A ; 2 2

_____________________ Slide 144 ____________________ §

δ = ( i + e – A). δ m = ( 2i– A).

§

111

Now, the refractive index for thin prism is given by δ + A  δmin + A sin  min 2   2 µ= = A A  sin   2 2 or, δmin = (µ − 1)A. _____________________ Slide 145 ____________________

Illustration A rectangular block of refractive index µ is placed on a printed page lying on a horizontal surface as shown in the figure. Find the minimum value of µ so that the letter L on the page is not visible from any of the vertical sides.

δ + A  sin  min   2  A  sin   2

_____________________ Slide 146 ____________________ Illustration _____________________ Slide 142 ____________________ Graphical representation of minimum deviation §

The graph shows the minimum deviation produced by the prism.

A ray of light PQ is incident at an angle i on face ML of a prism and is refracted along OR (figure). This ray, after refraction at face MN, travels along RN at grazing emergence. If µ is the refractive index and A refracting angle of the prism, show that 2   sini + cosA   µ = 1 +    sinA    

1

2

_____________________ Slide 147 ____________________ Dispersion _____________________ Slide 143 ____________________

§ §

Thin prisms § §

In thin prisms, the distance between the refracting surfaces is negligible and the angle of prism (A) is very small. For thin prism, δm is very small. i.e.,

§ §

It explains the blue colour of the sky at day time. When while light passes through a prism it is spilled into its seven constituent colours which is known as dispersion of white light. It is because for different constituent colours of light there are different values of Refractive index (µ). The violet ray bends maximum while red at least.

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112

Ray Optics & Optical Instruments Slide 148

§

We get a continuous bends of seven colours on screen, which is called spectrum of light as shown in the fig.

Physics Slide 153

Mathematical expression of Refraction at spherical surface §

Two media of refractive index µ1 and µ2, when separated by a transparent curved surface, can be regarded as the case of refraction at spherical surface.

_____________________ Slide 149 ____________________ Scattering of light § §

When sunlight travels through the earth’s atmosphere, it gets scattered by the atmospheric particles. The amount of scattering is given by Rayleigh scattering. _____________________ Slide 150 ____________________

Refraction at a spherical surface separating two media _____________________ Slide 154 ____________________

Rayleigh scattering §

§

The amount of scattering S is inversely proportional to the fourth power of the wavelength (λ). 1 S= 4 λ The law is applicable only for small particles which have relative size with the wavelength of the light. _____________________ Slide 151 ____________________

§ § § § § §

The figure shows how light refracts at the interface of two curved media. C is the centre of curvature of medium ‘2’. CN is the normal to the curved surface and O is the point where the object lies. After refraction, let the image be formed at I. Let ‘u’ be the object distance, and ’v’ be the image distance.

Spherical surface § §

A spherical refracting surface is a part of a sphere of refracting material. It is of two types.

_____________________ Slide 155 ____________________ §

From figure, for small angles, tan q ; q tan ÐNOM =

Convex refracting surface §

A refracting surface which is convex towards the medium where the object is present.

and

_____________________ Slide 152 ____________________

tan Ð NIM =

Refraction at spherical surface: Assumptions § § §

Object is a point object lying on the principle axis. The incident and the refracted rays make small angles with the principal axis. The aperture is small.

tan Ð NCM =

§

MN ; Ð NOM OM

MN ; Ð NCM MC MN ; Ð NIM MI

From plane geometry i=

MN MN + OM MC

− − − (i )

(exterior angle = sum of interior angles ; i = ∠NOM + ∠NCM)

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Physics

Ray Optics & Optical Instruments Slide 161

Slide 156 §

Similarly,

r=

MN MN – MC MI

− − − (ii )

(Q r = ∠ NCM – ∠ NIC) § §

Now from Snell’s law, µ1 sin i = µ2 sin r and for small angles sinθ ≈ θ. Hence, µ1 i = µ 2 r

§ §

From (i) and (ii) µ1 µ 2 µ 2 – µ 1 + = − − − ( iii) OM MI MC By using sign convention, OM = – u, MI = + v, MC= + R Hence, from eqn. (iii) µ2 µ1 µ2 – µ1 – = v u R _____________________ Slide 158 ____________________

Important terms § § § §

The medium in which the incident ray travels is called the 1st medium (µ1). The medium in which the refracted ray travels is called the 2nd medium (µ2). The above equation relates object and image distance in terms of refractive index and radius of curvature. This formula is applicable for both convex and concave surfaces with proper sign conventions. _____________________ Slide 159 ____________________

∴ v = 270 cm As v is positive, the image is formed in the second medium at a distance of 270 cm from the refracting surface. The image is real. _____________________ Slide 162 ____________________ Illustration An empty spherical flask of diameter 30 cm is Placed in water of refractive index. A parallel beam of light strikes the flask. Where does it get focused, when observed from within the flask? _____________________ Slide 163 ____________________ Illustration What curvature must be given to the bounding surface of a refracting medium (µ = 1.5) for the virtual image of an object in the adjacent medium (µ = 1) at 10 cm to be formed at a distance of 40 cm? _____________________ Slide 164 ____________________ Refraction by a lens Lens § §

Solved example A convex refracting surface of radius of curvature 15 cm separates two media of refractive indices 4/3 and 1.5. An object is kept in the first medium at a distance of 240 cm from the refracting surface. Calculate the position of the image.

4 / 3 1.5 1.5– ( 4 / 3 ) + = 240 v 15

∴

_____________________ Slide 157 ____________________ §

113

A combination of two refracting surfaces, at least one of which is curved, is called a lens. There are two types of lens ü Concave lens ü Convex lens. _____________________ Slide 165 ____________________

Concave lens §

Concave lens is a lens which is thinner at middle and wider at its ends and diverge light rays.

§

Concave lens is of three types

_____________________ Slide 160 ____________________ Solution As the object is in the rarer medium (i.e. the incident ray travels in the rarer medium), we have µ1 µ2 µ2 – µ1 + = –u v R Here, u = – 240 cm: v = ? ; R = + 15 cm, µ1 = 4/3 ; µ2 = 1.5

_____________________ Slide 166 ____________________ §

A concave lens can be somewhat supposed to be made up of two prisms placed on each other, vertex to vertex as shown in figure

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114 §

Ray Optics & Optical Instruments

Concave lens is also called a diverging lens because it diverges parallel beam of light after refraction through it.

Physics Slide 169

_____________________ Slide 170 ____________________

_____________________ Slide 167 ____________________ Convex lens §

Convex lens is a lens which is thicker at middle and thinner at its ends and converges light rays.

§

Convex lens is of three types

_____________________ Slide 171 ____________________ _____________________ Slide 168 ____________________ §

A convex lens is supposed to be made up of two prisms placed on each other, base in to base contact, as shown in fig.

§

Convex lens is also called converging lens because it converges parallel beam of light after refraction through it as shown in fig.

Note In general, focal length of a concave lens is taken negative and that of convex lens is taken positive. _____________________ Slide 172 ____________________

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Physics

Ray Optics & Optical Instruments Slide 173

115 Slide 177

§ § §

The geometry of image formation by a double convex lens. The first refracting surface forms the image I1 of the object O. The image I1 acts as a virtual object for the second surface that forms the image at I. _____________________ Slide 178 ____________________

Applying to the first interface ABC, we get §

To the second interface ADC, n2 n1 n – n1 – + =2 DI1 DI DC2

§

For a thin lens, BI1 = DI1.

_____________________ Slide 174 ____________________ Lens maker’s formula

n1 n n – n1 + 2 = 2 OB BI1 BC1

_____________________ Slide 179 ____________________ §

Adding and, we get æ 1 n1 n 1 ö÷÷ + 1 = ( n2 – n1) ççç + çè BC1 DC2 ÷÷ø OB DI1

§

Suppose the object is at infinity, i.e,, OB →∞ and DI1 = f, gives By the sign convention, BC1 = + R 1, DC2 = – R2 So, it can be written as æ1 1 1ö æ n ö = (n21 –1) ççç – ÷÷÷ çççQ n21 = 2 ÷÷÷ f n1 ø÷ èç R1 R2 ø÷ èç

§ §

_____________________ Slide 175 ____________________ §

Where n21 is the refraction index of the medium 2 w.r.t medium 1. Equation is known as the lens maker’s formula. _____________________ Slide 180 ____________________

Important Points § § § § _____________________ Slide 176 ____________________

Len’s maker’s formula is used to find the lens of desired focal length. While using this formula we should use proper sign conventions. A lens has two focii, two radius of curvature, one optical centre (same function as that of pole in mirror) The focus on the side of the original source of light is called the 1st focal point, while the other is called 2nd focal point. _____________________ Slide 181 ____________________

Relation between object distance (u), Image distance (v) and focal length, (f) of a thin lens §

The formula for object distance(u), Image distance (v) and focal length (f) of a lens is given by 1 1 1 – = v u f

§

This formula is applicable for both convex and concave lenses with proper sign convention.

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Ray Optics & Optical Instruments Slide 182

Rules for image formation through a lens § § § §

To find out the position of image through a lens we should find out the point of intersection of refracted rays. A ray parallel to principal axis of a lens after refraction passes through or appears to diverge from the focus of lens. A ray through optical centre of lens passes undedicated along its original path after refraction through lens. A ray through focus of lens becomes parallel to principal axis, after passing through the lens. _____________________ Slide 183 ____________________

§ § §

Physics

m is negative for real image which are inverted w.r.t. to object. m is positive for virtual image which are erect w.r.t. object. Mathematically, magnification is given by v f–v f  1 1 1 m= = = using, – =  u f f + u  v u f _____________________ Slide 186 ____________________

Solved Example A convex lens is to be used to throw on a screen 10 cm from the lens, a magnified image of an object. If the magnification is to be 19, find the focal length of the lens. _____________________ Slide 187 ____________________

Image formation by a concave lens

Solution Given: magnification, m = −19, Image distance, v = 10 cm Now, magnification is given by f–v m= , f f–10 or, –19 = f or, f = 0.5 cm _____________________ Slide 188 ____________________ _____________________ Slide 184 ____________________

Power of lens § §

The degree of converging or diverging of light rays through a lens is called its power. The power of a lens is the reciprocal of the focal length of the lens. where f is measured in metre. _____________________ Slide 189 ____________________

§

§ §

The power of a lens is the tangent of the angle by which it converges or diverges a beam of light falling at unit distance from optical centre. The SI unit of power of lens is called Dioptre (D). The power of convex lens is positive and that of concave is negative. _____________________ Slide 190 ____________________

Combination of thin lenses in contact

_____________________ Slide 185 ____________________ Magnification § §

Magnification (m) produced by a lens is defined as the ratio of the size of the image to that of the object. It is denoted by m.

Image formation by a combination of two thin lenses in contact

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Physics

Ray Optics & Optical Instruments Slide 191

§ §

§

Consider two thin convex lenses A and B having focal length f1 and f2 are placed in contact with each other as shown in fig. Image formed by the 1st lens ‘A’. is given by 1 1 1 – = − − − (i ) v1 u f1 For image formed by second lens B, we get 1 1 1 – = - - - (i i ) v v1 f 2 _____________________ Slide 192 ____________________

§

§

§

Adding (i) and (ii) we get 1 1 1 1 – = + v u f1 f2 If the two lens system is regarded as a single lens, then the equivalent focal length fe is related as 1 1 1 – = v u fe 1 1 1 Hence, = + f e f1 f2

117 Slide 196

Primary rainbow § §

The primary rainbow has violet colour on the inner edge and the red colour on the outer edge of the rainbow. Primary rainbow is formed due to two refractions and one total internal reflection of the light incident on the droplets. _____________________ Slide 197 ____________________

Secondary rainbow § § §

It has red colour on the inner edge and violet colour on the outer edge of the rainbow. It is formed due to refraction and two total internal reflections of light incident on the droplets. Order of colour of the secondary rainbow is just reverse of that of the primary rainbow. _____________________ Slide 198 ____________________

_____________________ Slide 193 ____________________ §

If several thin lenses are placed in contact then the effective focal length of their combination is given by 1 1 1 1 = + + +−−−−− f e f1 f2 f3

§

In terms of power, we can write P = P1 + P 2 + P3 +- - - - - - _____________________ Slide 194 ____________________

§

§

The total magnification m of the combination is a product of magnification of individual lenses m = m 1m 2 m 3 - - - - - - - - Such a system of combination of lenses is used in designing lenses for cameras, microscopes, telescopes and other optical instruments. _____________________ Slide 195 ____________________

The rainbow § § §

It is a beautiful patterns of colours seen in the sky after a shower The rainbow is an example of the dispersion of sunlight by the water drops in the atmosphere. The conditions for observing a rainbow are that the sun should be shining in one part of the sky while it is raining in the opposite part of the sky.

_____________________ Slide 199 ____________________ Scattering of light §

§

When s unlight travels through the earth’s atm osphere, its direction changes by the atm os pheric particles and this is called s cattering of light. The am ount of s cattering is invers ely proportional to the fourth power of the wavelength, called Rayleigh Scattering. Slide 200

§ §

Blue light being shorter wavelength, the sky look blue during daylight due to scattering. Red colour being least scattered light reaching our eyes, therefore, the sun looks reddish at sunrise and at sunset.

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Ray Optics & Optical Instruments Slide 201

Human eye

Physics Slide 204

Vision Defects

_____________________ Slide 202 ____________________ How the Eye Works:

Nearsightedness: (a) Light focuses before the retina Cannot see distant objects Gets worse as the body grows (b) Fix: concave lens which will focus the light back on the retina _____________________ Slide 205 ____________________ Vision Defects (cont) Farsightedness:

_____________________ Slide 203 ____________________

(c) Light focuses behind the retina Cannot see close objects Gets worse as the body ages (40+): lens is less flexible (d) Fix: convex lens which will focus the light back on the retina _____________________ Slide 206 ____________________ Contact Lenses § § §

Rests on a layer of tears between it and the cornea Produces the same result as eyeglasses Most refraction occurs at air-lens surface where change in refractive index is greatest.

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Ray Optics & Optical Instruments

119 Slide 211

Defects of vision Short sightedness (or myopia) § § §

A person who can see the near objects clearly but cannot focus on distant objects is short sighted. This defect occurs if a person’s eyeball is larger than the usual diameter. In such a case, the image of a distant object is formed in front of the retina as shown in fig.

_____________________ Slide 207 ____________________ § § § §

The light from an object enters the eye through the transparent cornea. It passes through a transparent lens held in place by ciliary muscles and is focused on the retina. The retina is sensitive to light and sends messages to the brain by way of optic nerve. An iris in front of the lens changes in size to regulate the amount of light that enters the eye.

_____________________ Slide 212 ____________________ Correction To correct short-sighted vision, a diverging lens (concave lens) of suitable focal length is placed in front of the eye.

_____________________ Slide 208 ____________________ § § §

§

It may be noted that the image of the object formed on the retina is real and inverted w.r.t. object. However, our brain interprets it as an erect image w.r.t. object. The human eye is about 2.5 cm in diameter and its neat spherical shape is maintained by the pressure of the fluid within it. A normal eye can focus on object located anywhere from about 25 cm to hundreds of kilometers away. _____________________ Slide 209 ____________________ _____________________ Slide 213 ____________________

Power of accommodation of the eye §

The ability of the eye to focus on objects at different distances is called power of accommodation of the eye.

Far point of the eye §

The most distant point that the eye can see clearly is called the far point of the eye. _____________________ Slide 210 ____________________

Near point of the eye §

§

Long sightedness (or hypermetropia) § § §

A person who can see distant objects clearly but cannot focus on near objects clearly is farsighted. This defect may occur if the diameter of person’s eyeball is smaller than usual. In such a case, for an object placed at the normal near point (i.e., 25 cm from eye), the image of the object is formed behind the retina as shown in fig.

The closest point at which an object is seen most clearly without strain is called the near point of the eye/ least distance of distinct vision (D) For an adult with normal eye, this distance is taken to be 25 cm (by convention). SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

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Ray Optics & Optical Instruments Slide 214

Physics Slide 218

Correction

Solution

To correct for long sighted vision a converging lens of suitable focal length is placed in front of the eye.

Since the far point of the short-sighted person is 100 cm away, he can see objects situated up to a distance of 100 cm. so, the lens used should be such that it forms the image (virtual) of the distant object at a distance of 100 cm. ∴ u = – ∞, v = – 100 cm _____________________ Slide 219 ____________________ Now, ∴ or, or,

1 1 1 = – f v u 1 1 1 = – f –100 – ∞ 1 1 = f –100 f = –100 cm

So, a concave lens of 100 cm focal length should be used. _____________________ Slide 220 ____________________ _____________________ Slide 215 ____________________ Presbyopia § § § §

It is the defect in which a person can see nearby objects due to the increase of the near point of the eye. In this, the near point increases to 200 m. It is caused due to the decreasing effectiveness of the ciliary muscle and the loss of flexibility of the lens. It is corrected by using a converging lens for reading. _____________________ Slide 216 ____________________

Astigmatism §

§ §

It is a defect in which eye lens looks at a wire eye mesh or a grid of lines, focusing in either the vertical or the horizontal plane which may not be as sharp as in the other plane. ‘It is caused due to the non-uniformity in the spherical shape of the cornea. It can be corrected using a cylindrical lens of desired radius of curvature with an appropriately directed axis.

Illustration 1 Where an object should be placed from a converging lens of focal length 20 cm so to obtain a real image of magnification 2? _____________________ Slide 221 ____________________ Photometry §

Important terms Luminous flux § It is the amount of light energy emitted per second by a source. § The unit of luminous flux is lumen § One lumen is the luminous flux emitted per unit solid angle by a uniform point source of luminous intensity 1 candela. _____________________ Slide 222 ____________________ Luminous intensity §

Note §

Astigmatism can occur along with myopia or hypermetropia. _____________________ Slide 217 ____________________

Solved Example Far point of a short-sighted person is 100 cm. What lens should he use to see distant objects clearly?

Photometry is the science of measurement of light in terms of its perceived brightness to the human eye.

§ §

Luminous intensity of source in a given direction is defined as the luminous flux per unit solid angle in that direction. The unit of luminous intensity is candela (cd). Candela is defined as the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 5.40 × 1014 Hz and that has a radiant intens ity in the direction of watt per steradian.

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Ray Optics & Optical Instruments

121 Slide 227

Magnifying power of a simple microscope The angular magnification or magnifying power of a simple microscope is the ratio of the angle subtended at the eye by the image at the near point and the angle subtended at the unaided eye by the object at the near point. _____________________ Slide 228 ____________________ The linear magnification is given by m =

§ _____________________ Slide 223 ____________________ or,

Illuminance or Intensity of illumination §

§ §

Illuminance or Intensity of illumination of a surface at any point may be defined as the luminous flux incident normally on a unit area of the surface held at that point. It is generally denoted by E. If luminous flux ∆φ falls normally on an area ∆A of a surface, then illuminance is given by

or,

§

The unit of illuminance is lux. Lux or metre candle is the illuminance produced at the inner surface of a sphere of one metre radius when a source of one candela is placed at its centre. Radiant flux is measured in watt.

§

§ §

 v m = 1–   f Here, in accordance with sign Convention v = − D[ far point] D ∴ m = 1+ f _____________________ Slide 229 ____________________

Magnification when the image is at infinity §

_____________________ Slide 225 ____________________ Luminous efficiency

1 m=v  u  1 1 m=v –  v f 

or,

_____________________ Slide 224 ____________________ § §

v u

§

The luminous efficiency of a light source is the ratio of the luminous flux emitted and the input electrical power. The unit of luminous efficiency is lumen/watt.

Consider an object having height h when it is at the near point D of the eye. h ∴ tan θ0 = D Since θ0 is small h ; q0 ... (1) D _____________________ Slide 230 ____________________

_____________________ Slide 226 ____________________ Simple Microscope It is an optical instrument used to see small and minute particles. §

Now, magnification when the object is at infinity. h v m= I = h0 u

or,

hI =

v h u 0

.... (2)

Where h I = image distance, h 0 = object distance. _____________________ Slide 231 ____________________ §

The angle subtended by the image h tan qi = I –v h v é \ using eqN (2 )ù = 0. êë úû –v u

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122

Ray Optics & Optical Instruments =–

h0 u

–h0 ; qi u

\

Slide 235 é\ forsmall qi ê êtan q ; q i i ë

ù ú ú û

The angle subtended by the object, when u = −f is given h0 by ; qi f

§

Physics

Telescopes § §

Light collectors Two types: ü Reflectors mirrors) ü Refractors (Lenses)

_____________________ Slide 232 ____________________ §

Now, the angular magnification is given by q m= i q0 =

\

m=

h0 D ´ f h0 D f

_____________________ Slide 233 ____________________ §

So, smaller the focal length of the lens, greater will be the magnifying power.

§

The simple microscope may be used in such a way that the image is formed at infinity.

Note §

_____________________ Slide 236 ____________________ Yerkes 40” Refractor- The World’s Largest

The maximum angular magnification is obtained when the image is the near point.

§

The minimum angular magnification is obtained when the image is at infinity. _____________________ Slide 234 ____________________

Light and Telescopes

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Physics

Ray Optics & Optical Instruments Slide 240

Slide 237 Reflecting Telescopes

123

§

§

The Hubble Space Telescope objective suffers from this (it is too flat by 2 microns, about 1/50 the width of a human hair) so it uses corrective optics to compensate. The corrective optics intercept the light beams from the secondary mirror before they reach the cameras and spectrographs. Fortunately, the Hubble Space Telescope's spherical aberration is so perfect, that it is easy to correct for! _____________________ Slide 241 ____________________

_____________________ Slide 238 ____________________ Before Costar

Reflection § §

Light that hits a mirror is reflected at the same angle it was incident from Proper design of a mirror (the shape of a parabola) can focus all rays incident on the mirror to a single place

After Costar

_____________________ Slide 242 ____________________ Famous Telescopes: Galileo § §

First telescope: 3x magnification Last one: 32 x

_____________________ Slide 243 ____________________ Famous Telescopes: Newton § § §

First reflector ever Built around 1670 After this: gargantuan telescopes!

_____________________ Slide 239 ____________________ Whoops - The Hubble needs a contact lens §

§

Both types of telescope can suffer from a defect called spherical aberration so that not all of the light is focused to the same point. This can happen if the mirror is not curved enough (shaped like part of a sphere instead of a paraboloid) or the glass lens is not shaped correctly. SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

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Ray Optics & Optical Instruments Slide 244

Physics Slide 248

Famous Telescopes: Lord Ross

_____________________ Slide 249 ____________________

_____________________ Slide 245 ____________________ Famous Telescopes: Mt Palomar

Telescope Size § §

A larger telescope gathers more light (more collecting area). Angular resolution is limited by diffraction of light waves; this also improves with larger telescope size _____________________ Slide 250 ____________________

5 Meter Telescope – Huge and heavy mirror _____________________ Slide 246 ____________________ Famous Telescopes: Arecibo Radio Telescope

_____________________ Slide 247 ____________________

_____________________ Slide 251 ____________________ Largest Earth-Based Telescopes §

Hobby-Eberly Telescope, Davis Mountains, TX ü 11 m diameter ü

§

Cannot see all parts of the sky

Keck I and II, Mauna Kea, HI ü 36 ×1.8 m hexagonal mirrors; equivalent to 10 m ü

Above most of atmosphere (almost 14,000 ft ASL)

ü

Operating since 1993

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_____________________ Slide 254 ____________________

_____________________ Slide 252 ____________________ Resolving Power of Telescopes

_____________________ Slide 255 ____________________ Andromeda Galaxy Telescope 1

Telescope 2 of double size

Not to be outdone, the European Union has proposed building the 100 meter OWL (Overwhelmingly Large Telescope)

_____________________ Slide 253 ____________________ Resolving Power of Telescopes (II) Andromeda Galaxy Resolution: (a) 10’ (c) 5”

(b) 1’ (d) 1”

_____________________ Slide 256 ____________________ Compound Microscope § §

A compound microscope is one which has much larger magnification to see very small objects. It uses two lenses one compounding the effect of the other.

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Ray Optics & Optical Instruments Slide 257

§

Physics

It is clear from the above equation that the smaller the focal length of the objective and eyepiece, larger is the magnifying power.

Magnifying power of a compound microscope

_____________________ Slide 262 ____________________ 1 1 1 – = vo uo fo

Again, or , or, or, ∴

_____________________ Slide 258 ____________________ §

Angular magnification or magnifying power of a compound microscope is defined as the ratio of the angle subtended by the final image at the eye to the angle subtended by the object seen directly, when both are placed at the least distance of distinct vision. _____________________ Slide 259 ____________________

§

§

The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective. M = Me × Mo Where Me and Mo are the magnifying powers of the eyepiece and objective respectively. D Now, M e = 1 + fe

vo vo vo – = v o uo fo vo v – = –1 + o uo fo vo v = 1– o uo fo  v  D M =  1– o   1 +  f f  o  e 

_____________________ Slide 263 ____________________ §

When the final image is formed at infinity, then L D M = Mo × Me = – × fo fe

§

In this case, the microscope is said to be in normal adjustment. _____________________ Slide 264 ____________________

Historical Optical Microscopes

Where fe is the focal length of the eyepiece. _____________________ Slide 260 ____________________ §

§

Also, Mo =

vo , Where vo is the distance A’B’ from the uo

objective and uo is the distance of the object from the objective. v  D Now, the magnification is given by M = o  1+  uo  f e 

_____________________ Slide 265 ____________________

_____________________ Slide 261 ____________________ § § §

Since the object is placed very close to the principle focus of the objective therefore u o is nearly equal to fe. i.e. u 0 ≈ f 0 vo is nearly equal to the length L of the microscope tube. i.e. v0 ≈ L L is the separation between the two lenses. M= –

L D  1+  fo  fe 

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Ray Optics & Optical Instruments

127 Slide 269

Slide 266 Solved Example

A person with a normal near point (25 cm) using compound microscope with an objective of focal length 8.0 mm and an eyepiece of focal length of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? What is the magnifying power of the microscope? _____________________ Slide 270 ____________________ Solution Given:

_____________________ Slide 267 ____________________

D = 25 cm; fo = 0.80 cm ; u0 = – 0.9 cm

Objective: Using lens formula for the objective, we have, 1 1 1 – = v 0 u0 f0 1 1 1 or + = v 0 0.9 0.8 ∴ v0 = 7.2 cm Eyepiece: Using lens formula for the eyepiece, 1 1 1 we have, – = v e ue fe _____________________ Slide 271 ____________________ or

–

1 1 1 – = D u e fe

or

–

1 1 1 – = 25 u e 2.5

∴

ue = – 2.27 cm

(Q v e = D = – 2 5 c m )

Separation between the two lenses = v0 + |ue| = 7.2 + 2.27 = 9. 47 cm Magnifying power of microscope, _____________________ Slide 268 ____________________ Current Optical Microscopes

M = M0 × Me v  D  7.2  25  = 0 1 +  = 1+ = 88 u0  fe  0.9  2.5  _____________________ Slide 272 ____________________ Solved Example The total magnification produced by a compound microscope is 20. The magnification produced by the eyepiece alone is 5. The microscope is focused on a certain object. The distance between the objective and eyepiece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal lengths of the objective and eyepiece.

Upright

Inverted

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Physics Slide 277

Slide 273

Magnifying power of a telescope Magnifying power of a telescope m is the ratio of the angle β subtended the eye final image to the angle α which the object subtends at the lens or the eye. b h f f m» » . 0 = 0 a fe h fe Note The length of the telescope tube is (f0 + fe). _____________________ Slide 278 ____________________

_____________________ Slide 274 ____________________ Solution For eyepiece, Me = 5, u e =? ve = – D = – 20 cm v Me = e ue v –20 or ue = e = cm = –4cm Me 5 Now, fe = ? u e= – 4 cm, D = ve = – 20 cm 1 1 1 1 1 1 = – ; = – fe ve ue fe –20 –4

_____________________ Slide 279 ____________________

_____________________ Slide 275 ____________________

Types of Refracting telescope On simplification, fe = 5 cm For objective, M0 × Me = M or

M –20 M0 = = =–4 Me 5

Now, v0 + u e = 14 or v0 = 14 – ue = (14 – 4) cm = 10 cm Now,wehave v M0 = 1 – 0 f0 or, – 4 = 1–

10 f0

∴ f0 = 2cm _____________________ Slide 276 ____________________ Telescope § §

§ §

Astronomical telescope § §

These are the telescope which is used to observe celestial bodies like stars, comets etc beyond our earth. It produces an inverted image of the distance object.

Terrestrial telescope § §

These are the telescope which is used to observe objects on the earth. It provides an erect image of the distant object. _____________________ Slide 280 ____________________

Reflecting telescope These are modern telescope which use a concave mirror rather than a lens for the objective to observe distant objects.

These are optical instruments used to observe distant objects. It has an objective and an eyepiece, the objective has a large focal length and a much larger aperture than the eyepiece. Light from a distant object enters the objective and a real image is formed in the tube at its second focal point. The eyepiece magnifies this image producing a final inverted image. SAVANT EDUCATION GROUP E-17, East of Kailash, New Delhi – 110065. Ph.: +91-11-26224417 www.savantgroup.org

Physics

Ray Optics & Optical Instruments Slide 281

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CURRICULUM BASED WORKSHEET

Advantages of reflecting telescope § §

It is free from spherical and chromatic aberration. Mirrors used in the reflecting telescope are easier to manufacture, cheaper than lens, lighter than an equivalent.

§

The light gathering power of a reflecting type telescope is so large that the final image formed is brighter. _____________________ Slide 282 ____________________

Disadvantage of reflecting telescope §

It is inconvenient to use because of frequent adjustments.

§

The objective mirror focuses light inside the telescope. _____________________ Slide 283 ____________________

Solved Example A refracting telescope has an objective of focal length 30 cm and an eyepiece of focal length 3 cm. It is focused on a scale distant 2 m. For seeing with relaxed eye calculate the separation between the objective and the eyepiece. _____________________ Slide 284 ____________________ Solution For seeing with relaxed eye, the final image should be formed at infinity. This is possible if the image formed by objective is at the focus of the eyepiece. Let v0 be the distance of the image formed by the objective. Using lens formula for the objective lens, we have, 1 1 1 + = v 0 200 30 or, v0 = 35.3 cm ∴ Separation between the objective and the eyepiece = v0 + f e = 35.3 + 3 = 38. 3 cm _____________________ Slide 285 ____________________ Resolving power of an optical instrument §

The ability of an optical instrument to produce distinctly separate images of two objects very close together is called resolving power of the instrument.

§

It depends on the diameter of the objective lens.

Topics for Worksheet – I Reflection of light on plane and curved surface Worksheet – I 1.

A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of radius of curvature 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved? 2. An object is placed in front of a convex mirror of focal length 60 m. If image formed is half of its size, find the position of image. 3. A square wire of side 3.0 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What is the area enclosed by the image of the wire? Given: the centre of the wire is on the axis of the mirror, with its two sides normal to the axis. 4. An object is placed at a distance of 36 cm from a convex mirror. A plane mirror is placed in between, so that the two virtual images so formed coincide. If the plane mirror is at a distance of 24 cm from the object, find the radius of curvature of the convex mirror. 5. A man 2 m tall, whose eye level is 1.84 m above the ground, looks at his image in a vertical mirror. What is the minimum vertical length of the mirror if the man is to be able to see the whole of himself? 6. Find the position of an object which when placed in front of a concave mirror of focal length 20 cm produces a virtual image which is twice the s ize of the object. 7. An object is placed (i) 10 cm (ii) 5 cm in front of a concave mirror of radius of curvature 15 cm. Calculate the position, nature and magnification of the image in each case. 8. An object is placed 15 cm from a convex mirror of radius of curvature 90 cm. Calculate image position and magnification. 9. A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and the magnification. Describe what happens as the needle is moved farther from the mirror. 10. An object is placed in front of a concave mirror of radius of curvature 40 cm at a distance of 10 cm. Find the position, nature and magnification of the image. 11. An object is placed at a distance of 40 cm from a concave mirror of focal length 15 cm. If the object is displaced through a distance of 20 cm towards the mirror, by how much distance is the image displaced.

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Physics

conclude about the minimum value of index of refraction of glass?

Refraction on a plane surface Worksheet – II A ray of light of frequency of 5 × 104 Hz is passes through a liquid. The wavelength of light measured inside the liquid is found to be 450 × 10−9m. Calculate the refractive index of the liquid. 2. A tank is filled with water to a height of 12.5 cm. The apparent depth of a meddle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water? If water is replaced by a liquid of refractive index 1.63 upto the same height, by what distance would the microscope have to be moved to focus on the needle again? 3. A small pin fixed on a table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table? Refractive index of glass = 1.5. Does the answer depend on the location of the slab? 4. Velocity of light in a liquid is 1.5 × 108 m s −1 and air, it is 3 × 108 m s −1. If a ray of light passes from this liquid into air, calculate the value of critical angle. *This explains the plane: Total internal Reflection. 5. A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out? Refractive index of water is 1.33. [Consider the bulb to be a point source of light.] 6. Velocity of light in glass in 2 × 108 m s −1 and that in air is 3 × 108 m s −1. By how much would an ink dot appear to be raised when covered by a glass plate 6 cm thick? 7. The bottom of a container is 4.0 cm thick glass (µ = 1.5) slab. The container contains two immiscible liquids A and B of depths 6.0 cm and 8.0 cm respectively. What is the apparent position of a scratch on the outer surface of the bottom of the glass slab when viewed through the container? Refractive indices of A and B are 1.4 and 1.4 respectively. 8. Refractive index of glass 1.5. Calculate velocity of light in glass if velocity of light in vacuum is 3 × 108 m s −1. Also calculate critical angle for glass-air interface. 9. Determine the critical angle for a glass-air surface, if a ray of light which is incident in air on the surface is deviated through 15°. When its angle of incidence is 40°. 10. Fig. shows a triangular prism of glass. A ray incident normally on one face is totally reflected. What can you 1.

11. (a) Fig. shows a cross-section of a light-pipe made of glass fibre of refractive index 1.68. The outer covering of the pipe is made of a material of refractive index 1.44. What is the range of the angles of incident easy with the axis of the pipe for which total reflections inside the pipe take place as shown in Fig. (b) What is the answer if the is no outer covering of the pipe?

Topics for Worksheet – III Prism Worksheet – III 1.

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Parallel light from the collimator of a spectrometer si incident on the two faces of a prism which make the refracting angle A of the prism. The image of the collimator slit is observed in two different positions of the telescope of the spectrometer. If the angle of rotation of the telescope between the two positions is 144°, what is the angle A of the prism? For a given source of light, the angle of minimum deviation of a 60° prism is 56°. What is its refractive index? If the refractive index of the material of a prism of refracting angle 8° is 1.532 for blue red light, what is the angular dispersion produced by the prism? A ray of light passes through an equilateral prism (µ = 1.5) such that angle of incidence is equal to angle of 3 emergence and the latter is equal to th of the angle 4 of prism. Calculate the angle of deviation. A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle of 3 emergence. The angle of emergence is times the 4 angle of prism. Calculate the refractive index of the glass prism.

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Physics 6.

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Ray Optics & Optical Instruments

Using a spectrometer, the following data are obtained for a crown glass prism and a flint glass prism. Angle of the prism, A = 72.0° Minimum deviation angle, δb = 54.6°, δr = 53.0°, δy = 54.0° Flint glass prism: A = 60.0 °, δb = 52.8′, δr = 50.6 ° δy = 51.9 ° b, r and y refer to particular wavelengths in the blue, red and yellow bands. Compare the dispersive powers of the two varieties of glass prisms. Find the angle of flint glass prism which produces the same angular dispersion for C and F wavelengths as a 10° crown glass prism. For crown glass: µF = 1.5230, µC = 1.5145 For flint glass: µF = 1.6637, µc = 1.6444. A small-angled prism (µ = 1.62) gives a deviation of 4.8°. Find the angle of the prism. The angle of minimum deviation for a prism of angle π π is . What is the velocity of light in the material of the 3 6 prism? Given: velocity of light in vacuum = 3 × 108 m s −1. Deduce δm for water (µ = 1.333) when the prism of ∠A = 60° is used. Calculate the dispersive power of crown glass if the deviations produced for violet, yellow and light are 4.32°, 4.02° and 3.72y respectively. Calculate the angle of dispersion between red and violet colours produced by a flint glass prism of refracting angle of 60°. Given: µv = 1.663 and µr = 1.622. A glass prism whose refractive index is 1.53 and refracting angle is 60° is held in a liquid of refractive index 1.33. Calculate the angle of minimum deviation in this case. Calculate the dispersive power for crown and flint glass from the following data: C D F Crown 1.5145 1.5170 1.5230 Flint 1.6444 1.6520 1.6637 A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a face of the prism. By rotating the prism, the minimum angle of deviation is measured to be 40°. What is the refractive index of the prism? If the prism is placed in water (refractive index 1.33), predict the new minimum angle of deviation of a parallel beam of light. The refracting angle of the prism is 60°.

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Topics for Worksheet – IV Refraction on a curved surface, Lens

13.

Worksheet – IV 1.

A beam of light travelling in air strikes a glass sphere of 20 cm diameter converting towards a point 40 cm behind the pole of the spherical surface. Find the position of the image, if the refractive index of glass is 1.5.

14.

131

The image obtained with a convex les in erect and its length is four times the length of the object. If the focal length of the lens is 20 cm, calculate the object distance and the image distance. The image of a needle placed 45 cm from a lens is formed on a screen placed 90 cm on the other side of lens. Find the displacement of image if the needle is moved 5 cm away from lens. Where should an object be placed from a converging lens of focal length 20 cm so to obtain a real image of magnification 2? An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved farther from the lens? A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm front P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, and (b) a concave lens of focal length 16 cm? The radius of curvature of each face of a bi-concave lens, made of glass of refractive index 1.5 is 30 cm. Calculate the focal length of the lens in air. Double convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length of the lens is to be 20 cm? A convex lens of focal length f and refractive index 1.5 is immersed in a liquid of refractive index (i) 1.6 (ii 1.3 (iii) 1.5. What is the new focal length in each case? (i) If f = + 0.5 m, what is the power of the lens? (ii) The radii of curvature of the faces of a double convex lens are 10 cm and 15 cm. Its focal length is 12 cm. What is the refractive index of glass? (iii) A convex lens has 20 cm focal length in air. What is the focal length in water? (Refractive index for airwater = 1.33, Refractive index for air-glass = 1.5.) Two thin converting lenses of focal lengths 0.15 m and held in contact with each other. Calculate power and focal length of combination. What curvature must be given to the bounding surface of a refracting medium (µ = 1.5) for the virtual image of an object in the adjacent medium (µ = 1) at 10 cm to be formed at a distance of 40 cm? An empty spherical flask of diameter flask of diameter 4 30 cm is placed in water of refractive index . A parallel 3 beam of light strikes the flask. Where does it get focused, when observed from within the flask. The image of a needle placed 10 cm from a lens is formed on a wall 20 cm on the other side of the lens.

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Find the focal length of lens and size of image formed if the size of object needle is 2.5 cm. A convex lens is to be used to throw on a screen 20 cm from the lens, a magnified image of an object. If the magnification is to be 19, find the focal length of the lens. A needle placed 45 cm from a lens forms an image on a screen placed 90 cm on the other side of the lens. Identify the type of lens and determine its focal length. What is the size of the image if the size of the needle is 5.0 cm? A converting beam of light passes through a diverting lens of focal length 0.2 m and comes to focus 0.3 m behind the lens. Find the position of the point at which the beam would converge in the absence of the lens. The radius of curvature of each surface of a convex lens of refractive index 1.5 is 0.40 m. Calculate its power. An equiconvex lens of focal length 15 cm is cut into two equal halves in thickness. What is the focal length of each half? A glass has a focal length of 5 cm in air. What will be its focal length in water? Refractive index of glass is 1.51 and that of water is 1.33. A glass convex lens has a focal length of 20 cm in air. What will be its focal length, when it is completely immersed in a liquid of refractive index 1.63? Given aµg = 1.5. (a) A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens. (b) Suppose the object in (a) above is in illuminated slit in a collimator tube so that it is hard to measure slit size and its distance from the screen. Using a convex lens, one obtains a sharp image of the slit on a screen. The image size is measured to be 4.6 cm. The lens is displaced away from the slit and at a certain location, another sharp image of size 1.7 cm is obtained. Determine the size of the slit.

Topics for Worksheet – V Optical instruments Worksheet – V 1. 2.

Far point of a short-sighted person is 100 cm. What lens should he use to see distance objects clearly? A short-sighted person can only see objects distinctly if they lie between 8 cm and 100 cm from the eye. What kind of lens should be required to see a star clearly and what would be its focal length? With these glasses, what would be the least distance of distinct vision?

Physics

A figure divided into squares, each of size 1 mm 2, is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to the eye. (i) What is the magnification (image size/object size) produced by the lens ? How much is the area of each square in the virtual image? (ii) What is the angular magnification (magnifying power) of the lens? (iii) Is the magnification in (i) equal to the magnifying power in (ii)? Explain. 4. A person with a normal near point (25 cm) using a compound microscope with and objective of focal length 8.0 mm and eyepiece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses? How much is the magnifying power of the microscope? 12. Two convex lenses of focal lengths 10 cm and 1 cm constitute a telescope. The telescope is focused on a scale which 1 m away from the objective. Calculate the magnification produced and the length of the tube if the final image is formed at a distance of 25 cm from the eye. 13. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope? (b) If this telescope is used to view the Moon, what is the diameter of the image of the Moon formed by the objective lens? The diameter of the Moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108 m. 14. (a) The image of the objective in the eyepiece is known as the ‘eye-ring’. Why is this the best position of our eyes for viewing? (b) Show that the angular magnification of a telescope equals the ratio of the diameter of objective to the diameter of eye-ring. (c) The angular magnification of a telescope is 300. What should be the diameter of the objective is our eyes (located at the eye-ring) are just able to collect all the light refracted by the objective? Take the diameter of the pupil of the eye to be 3 mm. 5. A person can see clearly only upto 3 metre. Prescribe a lens for his spectacles so that he can see clearly upto 12 metre. 6. A short-sighted person can see objects most distinctly at a distance of 16 cm. If he wears spectacles at a distance of 1 cm from the eye, what focal length should they have so as to enable him to see distinctly at a distance of 26 cm? 7. To print a photograph from a negative, the time of exposure to light from a lamp placed 0.50 m away is 2.5 second. How much exposure time is required if the lamp is placed 1.0 m away? 3.

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Physics

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8.

A convex lens of focal length 6.25 cm is used as a magnifying glass. If the near point of the observer is 25 cm from the eye and the lens is held close to the eye, calculate (i) the distance of the object from the lens and (ii) the angular magnification. 9. In the previous question, what should be the distance between the object ant the magnifying glass if the virtual image of each square in the figure is to have an area of 6.25 mm 2. Would you be to see the squares distinctly with your eyes very close to the magnifier? 10. An angular magnification (magnifying power) of 30 × is desired using an objective of focal length 1.25 cm and an eyepiece of focal length 5 cm in a compound microscope. What is the separation between objective and the eyepiece? 11. An amateur astronomer wishes to estimate roughly the size of the Sun using his crude telescope consisting of an objective lens of focal length 200 cm and an eyepiece of focal length 10 cm. By adjusting the distance of the eyepiece from the objective, he obtains an image of the Sun on the screen 40 cm behind the eyepiece. The diameter of the Sun’s image is measured to be 6.0 cm. What is his estimate of the Sun’s size, given that the average Earth-Sun distance is 1.5 × 1011 m? 12. A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. What is magnifying power of the telescope for viewing distance objects when (a) the telescope is in normal adjacent (i.e., when the final image is at infinity), (b) the final image is formed at the least distance of distinct vision (25 cm). (c) What is the separation between the objective and eye-lens in case (a) ? (d) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens? (e) What is the height of the final image of the tower if it is formed at 25 cm?

CURRICULUM BASED CHAPTER ASSIGNMENT

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1 Mark Questions 1.

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Radius of the curvature of convex mirror 40 cm and the size of the object is twice as that of the image. Then, what is the image distance? When a mirror is rotated through an angle θ, what is the angle turned by the reflected ray?. A concave mirror of focal length f produced an image n size of the object. If the image is real, what is the distance from the mirror? The angle between incident ray and the reflect ray is 70°. What if the angle of incidence?

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A 5 cm tall object is placed 10 cm from a concave mirror of focal length 15 cm. Find the position, nature and the size of the image. A tall man of height 6 feet want to see his image. Find the required minimum length of the mirror to see his image. Where should an object be placed in front of concave mirror focal length f so that the image to be the same size as the object? What is the magnification produced a plane mirror? If the power of a lens is +5 dioptres, what is the focal length? A parallel beam of light is incident on a concave lens of large aperture, Will the reflected coverage at point? Which of the two main parts of an optical fiber has a higher value of refractive index? A thin prism of 60° angle gives a deviation of 30°. What is the refractive index of material of prism? What is the refractive index of air for light waves? Mention two conditions of total internal reflection? What happens when white light passes through a prism? What is a prism? Show the refraction of light by a prism with a suitable diagram. Give two characteristic properties when the prism is in the position of minimum deviation? Which of the following colours suffers maximum deviation in a prism? What is the importance of linear magnification? Where should an object placed in image of the same by a convex lens? Can it happen in case of a concave lens? How will you judge whether a given piece of glass is convex lens, concave lens or a plane glass sheet? A lens whose radii of curvature are different is forming an image. If the lens is reversed, will the position of image change? What is the nature of the image formed by a concave mirror when the object is placed between its pole and focus? Focal length of an equiconvex lens is equal to the radius of curvature of either face. What is the refractive index of lens material? What should be the position of an object relative to a biconvex lens so that it behaves like a magnifying lens? A converging and a diverging lens of equal focal lengths are place coaxially in contact. Find the power and focal length of the combination. When a biconvex lens made of glass (µ = 1.5) is immersed in water (µ = 1.33), what will happen to its focal length?

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2 Marks Que stions 29. What happen when a monochromic ray passes through a prism? 30. State the law of reflection. 31. What do you mean by the spherical aberration of a lens? How can it be minimized? 32. Define linear magnification? 33. Mention the use of plane, concave and the convex mirror. 34. An object is placed between the focus and the pole of a concave mirror. Locate the position of the image by the ray diagram. 35. What is mean by the linear magnification? 36. Give new Cartesian sigh convention for the spherical mirrors. 37. A ray of light incident on one face of an equilateral prism undergoes total internal reflection at another face. If the refractive index of prism material is 2 then, find the angle of incidence (r2) at another face is 38. A ray is incident at an angle of incidence i on one surface of a prism of small angle A and emerges normally from the opposite face. If the refractive index of the prism material is µ, then what angle of incidence? 39. If the angle of a prism is 60° and angle of minimum deviation is 40°. Then find the angle of refraction. 40. Refractive index of glass for light of yellow, green and red colours are µy, µg and µr respectively. Rearrange these symbols in an increasing order of values? 41. At what angle of incidence should a light beam strike a glass slab of refractive-index 3 , such that the reflected and the refracted rays are perpendicular to each other? 42. Show with the help of a diagram the deviation produced by a prism. 43. Define a monochromatic and polychromatic light. How can one obtain monochromatic light from polychromatic light? 44. How deep will a 4 m deep tank appear when seen in air due to optical illusion? Refractive index of water is 4/3. 45. What is dispersion? 46. A concave lens of f = 15 cm forms an image 10 cm from the lens. Prove that the object is placed 30 cm away from the lens. 47. An object is placed 10 cm in front of a lens. The lens forms a real image three times magnified. Where is the image formed? What is the focal length of the lens? 48. The radius of curvature of either face of a convex lens is equal to its focal length. What is the refractive index of the material of the lens? 49. Calculate the focal length of the combination of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm. Is the system a converging or a diverging lens?

Physics

50. A compound microscope with an objective of 1.0 cm focal length and an eye-piece of 2.0 cm focal length has a tude length of 20 cm. calculate the magnifying power of the microscope, if the final image is formed at the neat of the eye. 51. An astronomical telescope, in normal adjustment position has magnifying power 5. The distance between the objective and the eye-piece is 120 cm. calculate the focal lengths of the objective and of the eye-piece. 52. A convex lens of refractive index n1 is held in a medium of refractive index n2 . trace the path of refracted rays of parallel beam of light incident on the lens when (i) n 2>n1 and (ii) n 2=n1. 53. Using the lens formula, show that a concave lens produces a virtual and diminished image independent of the location of object.

3 Marks Questions 54. An object is placed 10 cm in front of a concave mirror of focal length 15 cm. Find the nature, position and size of the image. 55. An erect image three times the size of the object is obtained with the concave mirror of radius of curvature 36 cm. What is the position of the object? 0

56. Light of wavelength 5000 A fall on a reflection surface. What are the wave length and frequency of reflected ray normal the incident ray? 57. What type mirror will you prefer for the saving or make –up? 58. For dividing a car, which type of mirror would you prefer to see the traffic at your back? 59. What is the difference between virtual image formed by the plane, concave and convex mirror? 60. Described a simple method of finding the focal length of a concave mirror. 61. What are spherical aberration? 62. What do you understand by diffuse reflection? 63. Draw three important rays for a concave mirror that are generally used to locate the position of the image of an object? 64. Define the dispersive power of a prism. 65. Why does the sky appear blue? 66. A ray of light falls normally on the face of a prism of refractive index 1.5. Find the angle of the prism if the ray just fails to emerge from the second face. 67. The minimum deviation produced by a glass prism having an angle of 60° is 30°. If the velocity of light in air is 3 × 108 ms–1, calculate its velocity its velocity in glass. 68. A glass prism has a refracting angle of 60°. The angle of minimum deviation is 40°. Find the refractive index.

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Physics

Ray Optics & Optical Instruments

At what angle the ray of light should be incident so as to suffer minimum deviation? 69. A ray of light enters a rectangular glass slab of refractive index 3 at an angle of incidence 60°. It travels a distance 5 cm inside the slab and emerges out of the slab. What is the perpendicular distance between the incident and emergent rays? 70. Prove that the refractive index of denser medium w.r.t. rarer medium is equal to the reciprocal of the refractive index of rarer medium w.r.t. denser medium. 71. An object is immersed in water, show that Realdepth a µw = Apparent depth 72. Describe two applications of atmospheric refraction. 73. A myopic person has been using spectacles of power— 1.0 dioptre for distant vision. During old are he also needs to use separate reading glass of power +2.0 dioptre. Explain what may have happened. 74. A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at point O and PO = OQ. Find the distance of the object from the spherical surface. 75. A small point object is placed in air at a distance of 60 cm from the convex spherical refracting surface of refractive index 1.5. If the radius of the curvature of the spherical surface is 25 cm, find the position of the image and power of the refracting surface. 76. An ink dot marked on the surface of glass sphere placed in air is viewed Fig

through the glass from a position directly opposite. If the diameter of the sphere is 15 cm and the refractive index of glass is 1.5, find the position of the image.

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77. A convex refracting surface of radius of curvature 15 cm separates two media of refractive indices 4/3 and 1.5. An object is kept in the first medium at a distance of 240 cm from the refracting surface. Calculate the position of the image. 78. A real image of an object id formed at a distance 20cm from a lens. On putting another lens in contact with it, the image is shifted 10 cm toward the combination. Determine the power of the second lens. 79. How does the focal length of a convex lens change if monochromatic red light is used instead of monochromatic blue light? 80. Draw a ray diagram of an astronomical telescope in the normal adjustment position. Write down the expression for its magnifying power. 81. Draw a ray diagram of an astronomical telescope in the near adjustment position. Write down the expression for its magnifying power. 82. Explain why does a convex lens behave as a converging lens when immersed in water (µ = 1.33) and as a diverging lens, when immersed in carbon disulphide (µ = 1.6).

5 Marks Questions 83. Obtain the relation between radius of curvature (R) and the focal length (f) of a convex mirror? 84. A 2.0 cm tall is placed 15 cm from a concave mirror of focal length 10 cm. Find the position, size and nature of the image? 85. Two planes are mirror are inclined to each at angleθ. A ray of light is reflected first at one mirror and mirror then at the other. Fined the total deviation of the ray and show that it is independent of the angle of the incidence at first t mirror. 86. A square wire of side 3.0 cm is placed 25 cm away from a concave mirror of focal length is on the axis of the mirror with the two sides normal to the axis. 87. Two objects A and B when the placed in front of the concave mirror of focal length 7.5 cm give the image of equal size If A is three times the size from the mirror of B and is place 30cm, Find the distance of the object from the pole of the mirror, 88. The angle of minimum deviation for yellow light in a prism of refractive index 1.6 is found to be 46°. Calculate the refracting angle of the prism. 89. In a spectrometer, for the prism A = 60°, calculate the angle of minimum deviation if µ of the prism for orange light is 1.64? 90. Define critical angle, obtain a relation for it. Define total internal reflection. State essential conditions for its occurance. 91. A ray of light is incident at an angle of 60° on one face of a prism which has an angle of 30°. The ray emerging

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Ray Optics & Optical Instruments

92.

93.

94.

95.

96.

out of the prism makes an angle of 30° with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges and calculate the refractive index of the material of the prism. A ray of monochromatic light is incident on the refracting face of a prism of refracting angle 75°. It passes through the prism and is incident on the other face at the critical angle. If the refraction index of the prism is 2 , find the angle of incidence on the first face of the prism. A glass prism of angle 72° and refractive index 1.66 is immersed in a liquid of refractive index 1.33.Find the angle of minimum deviation for a parallel beam of light passing through the prism. An object is placed 20 cm to the left of a convex lens of focal length 10 cm. If a concave mirror of focal length 5 cm is placed 30 cm to right of the lens, what is the position of the final image? A diverging lens of focal length 20 cm is placed 60 cm from a pin and a concave mirror of radius of curvature 20cm is placed on the opposite side of the lens. Where should the mirror be placed so that image of the pin may coincide with the pin itself? With the help of the data shown in the ray diagram in fig, calculate the focal length of the concave lens.

6.

7.

8.

9.

One can not see through fog because: (a) Fog absorbs light (b) Light is scattered by the droplets in fog (c) Light surface total internal reflection by the droplets in fog (d) The refractive index of fog is infinity The refractive index of the medium if a light wave has a frequency of 4 × 1014 Hz and a wavelength of 5 × 10–7 m in a medium, will be: (a) 1.5 (b) 1.33 (c) 1.0 (d) 0.66 The number of wavelengths in the visible spectrum is: (a) 4000 (b) 6000 (c) 2000 (d) Infinite Find the length of the optical path of two media in contact of lengths d 1 and d2 of refractive indices µ1 and µ2 respectively. (a) µ1 d1 + µ 2d2 (b) µ1 d2 + µ 2d1 (b)

10.

d1d2 µ1µ2

2.

3.

4.

5.

d1 + d2 µ1µ2

3 3 and respectively: 2 4 9 (b) 8

water with respect to air are

8 9 7 (c) 6

QUESTION BANK FOR COMPETITIONS A ray is reflected in turn be three plane mirrors mutually at right angles to each other. The angle between the incident and the reflected rays is: (a) 90° (b) 60° (c) 180° (d) None of these A plane mirror is in front of you in which you can see your image. It is approaching towards you at a speed of 10 cm/s’ then at what speed will your image approach you? (a) 10 cm/s (b) 5 cm/s (c) 20 cm/s (d) 15 cm/s The size of the image, if an object of 2.5 m height is placed at a distance of 10 cm from a concave mirror is: (a) 10.5 m (b) 9.2 m (c) 7.5 m (d) 5.6 m Ray optics fails when the size of the obstacle is: (a) 5 cm (b) 3 cm (c) Less than the wavelength of light (d) (a) and (b) both. Arrange the following in ascending order of frequency. (a) Red, Blue, yellow, green (b) Blue, green ,yellow ,red (c) Red, yellow, green, blue (d) Red, green, yellow, blue

(d)

Calculate the refractive index of glass with respect to water. It is given that refractive indices of glass and

(a)

1.

Physics

11.

When a prism is dipped in water then the angle of minimum deviation of a prism with respect to air will be:

3 4   a µg = 2 ,a µw = 3    1 (a) 8 3 (c) 4 12.

13.

14.

(d) None of these

1 2 1 (d) 4 (b)

An astronomical telescope has a large aperture to: (a) Reduce spherical aberration (b) Have high resolution (c) Increase span of observation (d) Have low dispersion To get three images of a single object, one should have two plane mirrors at an angle of: (a) 60º (b) 90º (c) 120º (d) 30º A thin glass (refractive index 1.5) lens has optical power of – 5D in air. Its optical power in a liquid medium with refractive index 1.6 will be: (a) 1D (b) –1D (c) 25D (d) –25D

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Physics 15.

16.

17.

18.

Ray Optics & Optical Instruments

A plane mirror. (i) Can form real object (ii) Neither converges nor diverges the rays (iii) Cannot form real image of a real object. Choose the correct option or options. (a) (i) is correct (b) (i) and (ii) are correct (c) (ii) and (iii) are correct (d) None of all above A man of height 1.8 m stands in front of large vertical plane mirror. The distance of the image from the main if he stands at a dis tance of 1.5 m from the mirror is: (a) 1 m (b) 2 m (c) 3 m (d) 4 m A man is running towards a plane mirror with some velocity. If the relative velocity of his image with respect to him is 4 m/s, then the velocity of a man is: (a) 2 m/s (b) 4 m/s (c) 15 m/s (d) 16 m/s The mirrors are perpendicular to each other as shown in figure. A light ray a light ray AB is incident on the mirror M1. Then the reflected ray wills also surfer a reflection from the mirror M2. Then the final ray after reflection from M2 will be parallel to the incident ray, if: M1

20.

P

45° A

21.

22.

B

23.

19.

(a) I =45° (b) I = 65° (c) I < 30° (d) for any I between 0° and 90° A pole 5 m high on a horizontal surface. Sun rays rare incident at an angle 30° with vertical. The size or shadow on horizontal surface is: (a) 5 m (b) (c)

5 3 10 3

A point object P is situated in front of plane mirror shown in figure. The width of mirror AB is d. The visual region on a line passing through point P and parallel to the mirror is: d

i

M2

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B (a) d (b) 2d (c) 3d (d) None of the above A beautiful girl with two normal eyes wants to see full width of her face by a plane mirror. The eye to eye and ear to ear distances of her face are 4 inch and 6 inch respectively. The minimum width of the required mirror is: (a) 1 inch (b) 2 inch (c) 3 inch (d) 4 inch A ray f light falls on a plane mirror. When th e mirror is turned, about an axis at right angle to the plane of the mirror through 20°, the angle between the incident ray and new reflected ray is 45°.The angle between the incident ray and original reflected ray is: (a) 15° (b) 30° (c) 45° (d) 60° A lamp and scale arrangement, used to measure small deflected is shown in the figure. SS’ is the glass scale placed at a distance of 1 MM and I is the position of the light spot formed after reflection from the under deflected mirror MM. the mirror is deflected by 10° and comes to the deflected position MM. the distance moved by the spot on the scale is: M

10°

1m

m m

(d) None of these

S

M’

I

R M’

(a) 24.6 cm (c) 46.4 cm

S M

(b) 36.4 cm (d) 34.9 cm

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138 24.

25.

Ray Optics & Optical Instruments A plane mirror, reflecting a ray of incident light, is rotated through an angle θ about an axis through the point of incidence in the plane of the mirror perpendicular of the plane of incidence, then: (a) The reflected ray does not rotated (b) The reflected ray rotates through an angle θ (c) The reflected ray rotates through an angle θ (d) The incident ray is fixed In the given figure, the angle between reflected rays is equal to:

27.

Physics

Two plane mirrors are placed parallel to each other as shown in the figure. There is an object O placed between the mirrors, at 5 cm from mirror M2 . What are the distances of first three images from M2?

O

15 cm M1 A

28.

26.

(a) A (b) 2A (c) 3A (d) 4A A vessel consists of two plane mirrors at right angles (as shown in figure). The vessel is filled with water. The total deviation in incident ray is:

29.

30.

90°

(a) (b) (c) (d)

0° 90° 180° None of the above

M2

(a) 5,10,15 (b) 5,15,30 (c) 5, 25, 25 (d) 5,15,25 If two mirrors are inclined at some angle and an object is placed between the mirrors and there are 7 images formed for an object, then what is angle between the mirrors? (a) 54° (b) 50° (c) 60° (d) 45° If u represents object distance from pole of spherical mirror and v represents image distance from pole of mirror and f is the focal length of the mirror, then a straight line u = v measurement of f is: (a) (f,f) (b) (2f,2f) (c) (f, 2f) (d) (0,0) The position of 1 cm tall object which speed is placed 8 cm in front of a concave mirror of radius of curvature 24 cm is: (a) 24 cm (b) 25 cm (c) 26 cm (d) 27 cm

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