Teaching Notes 1

Name : Roll No. : Topic : Optics Mohammed Asif

Ph : 9391326657, 64606657

Teaching Notes (Optic)

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The study of light and vision is called optics. Light is a form of energy which is propagated as Electromagnetic waves which produces the sensation of sight in us.

Geometrical optics treats propagation of light in terms of rays and is valid only if wavelength of light much lesses than the size of obstacles. i) Light does not require a medium for its propagation ii) It’s speed in free space (vaccum) is 3 x 108m/s iii) It is transverse in nature In the spectrum of e.m.w. it lies between u.v. and infra-red region and has wavelength between O 4000 to 7000 A . i.e ( 0.4 µ m to 0.7 µ m )

Indigo is not distensible from blue. BASIC - DEFINATIONS • Source: A body which emits light is called source. Source can be a point one (or) extended one. (a) Self-luminous-source: The source which possess light of it own. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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(b)

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Ex:- Sun, Electric arc, Candle, etc. Non-luminous-Source: It is a source of light which does not possesses light of its own but acts as source of light by reflecting the light received by it. Ex: Moon, object around us, Book…….etc. Isotropic Source: It gives out light uniformly in all directions. Non-isotropic Source: It do not give out light uniformly in all direction. Medium: Substance through which light propagates is called medium

Ray: The straight line path along with the light travels in a homogeneous medium is called a ray. A single ray cannot be propagated form a source of light. Beam: A bundle can bunch of rays is called beam it is called beam it is of following 3 types Convergent-beam: In this case diameter of beam decreases in the direction of ray

Divergent Beam: It is a beam is with all the rays meet at a point when produced backward and the diameter of beam goes on increasing as the rays proceed forward.

Parallel Beam: It is beam in which all the rays constituting the beam move parallel to each other and diameter of beam remains same

Object: An optical object is decided by incident rays only. It is if two kinds Real Object: In this case incident rays are diverging and point of divergence is the position of real object.

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Virtual Object: In this case incident ray are converging and point of convergence is the position of virtual object. Virtual object cannot be seen by human eye be cause for an object can image to be seen by eyes, ray received by eyes must be diverging.

Image: An optical image is decided by reflected (or) refracted rays only. It is of two types. (a) Real Image: This is formed due to real intersection of reflected (or) refracted rays, Real image can be obtained on screen.

Virtual-Image: This is formed due to apparent intersection of reflected (or) refracted light rays. Virtual image can’t be obtained on screen. (Note: Human ray can’t distinguish between real and virtual image because in both case rays are diverging)

• REFLECTION:

The phenomenon by virtue of which incident light energy is partly or completely sent back into the same medium from which it is coming after being obstructed by a surface is called reflection. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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The direction of incident energy is called incident ray and the direction in which energy is thrown back is called reflected ray. It is of two types.

• LAWS OF REFLECTION:

1) First Law: The incident ray, the reflected ray and the normal to the reflecting surface at the point of incidence, all lie in one plane which is ⊥' r to the reflecting surface. 2) The angle of incidence is equal to the angle of reflection ∠i = ∠r . Note: 1) The laws of reflection are valid for any smooth reflecting surface irrespective of geometry.

2) Whenever reflection takes place, the component of incident ray parallel to reflecting surface remains uncharged, while component perpendicular to reflecting surface (i.e. along normal) reverse in direction.

→ ^ ^ ^

r1 = xi+ jy + zk

→ ^^^

,

r2 = xi+ jy − zk

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3) Vector form of laws of Reflection:

 R I−= 2 I.N N  ^ ^ ^^ ^

R → Unit vector along the reflected ray ^

I→ ^

N→ •

Unit vector along the Incident ray Unit vector along the normal ray

Image formed by a plane mirror: a) Point Source: For construction of image of a point source it is sufficient to consider any two rays falling on mirror. The point of intersection of corresponding reflected rays give the position of image as shown in figure.

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(∴∆ ABD

≅ ∆ABI

)

Image I lies as much behind the mirror as the object is in front of it. b) Extended source:

•

Characteristics of the image formed by a plane mirror: 1) The image formed by a plane mirror is Virtual 2) The image formed by a plane mirror is Erect 3) The image formed by a plane mirror is of same size as object. 4) The image formed by a plane mirror is at the same distance behind the mirror is the object is infront of it. 5) The image is laterally inverted (i.e.) right appear as left and vice-versa. 6) Note: If two plane mirror faring each other are inclined at an angle θ with each other, then number of images are formed due to multiple reflection. This principle is used in the toy kaleidoscope. (a) If

360

θ

is even integer, then number of images formed is η =

Ex: If θ = 60 0 then η = (b) If

•

360

θ

360 −1 = 6 −1 = 5 60

is odd integer, then number of images formed is η =

360

θ

−1

360

θ

360 =9 Ex: If θ = 40 0 (which is not the complete part of 1800) then η = 40

Deviation ( δ ): The angle between incident and reflected (or) refracted ray is termed as deviation. For reflection δ = π − 2 i

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Cases: When i = 0 (Normal incidence) δmax = π

When i =

π 2

Multiple Reflection:

(Grazing incidence) δ min = 0

δnet = ∑ δi

δi = deviation due to single reflection. Note while summing up, sense of rotation is taken into account.

Q: 1) Two plane mirror are inclined to each other such that a ray of light incident on the first mirror and parallels to the second is reflected from the second mirror parallel to the first mirror. Determine the angle between the two mirror. Also determine the total deviation produced in the incident ray due to the two reflections. Solution:

From figure

3θ =180

θ = 60 0 δ1 = π − 2 i =180 − 2 ×30 0 =120 ↑ A.C.W.

δ 2 = π − 2 i = 180 − 2 × 30 0 = 120 0

∴δnet = δ2 + δ1 = 240 ↑ ( or ) 120 ↓

Or From fig. δ = 180 + θ = 180 + 60 = 240 0 ↑

( or )

120 ↓

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Q: 2) Calculate deviation suffered by incident ray in situation as shown in figure, after three successive reflections? Solution: F,B.D

δ1 = π − 2 i =180 − 2 ×50 0 =100 ↑

δ 2 = 180 − 2 × 20 0 = 140 0 ↓ δ 2 = 180 − 2 ×10 0 = 160 0 ↓ δ net = 100 ↑ +140 ↓ +160 ↓ = 100 ↑ ( or )

260 0 ↓

Q: 3) Two plane mirrors are inclined to each other at an angle θ . A ray of light is reflected first at one mirror and then at the other. Find the total deviation of the ray? α = Angle of incidence for M1 Solution: Let β = Angle of incidence for M2 δ1 = Deviation due to M1 δ2 = Deviation due to M2

From figure

δ1 = π − 2 α δ2 = π − 2 β Also ray is rotated in same secure (i.e.) anticlockwise δ Net = δ1 +δ2 Now in ∆ OBC = π − 2α + π − 2 β ∠OBC + ∠BCO + ∠COB = 180 0 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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(90

δ Net = 2π − 2(α + β ) ∴ δ Net = 2π − 2(θ ) δ Net = 2π − 2θ

•

) (

)

− α + 90 0 − β + θ = 180 0 α + β =θ 0

Velocity of Image: Let xO/m = x-co-ordinate of object w.r.t. mirror xI/m = x-co-ordinate of image w.r.t. mirror yO/m = y-co-ordinate of object w.r.t. mirror yI/m = y-co-ordinate of image w.r.t. mirror For plane mirror

xO/m = -xI/m Differentiating both sides w.r.t. time (t) d ( xO / m ) = − d ( x I / m ) dt dt  →   →  VO / m  = −VI / m    x  x →

→

V Ox − Vmx = −VIx − V mx →

→

VIx = 2 V mx − V Ox Similarly yI/m = yO/m Differentiating both sides w.r.t. time we get  →   →  VI / m  = VO / m   y  y

In nutshell, for solving numerical problems involving calculation of velocity of image of object with respect to any observer, always calculate velocity of image first with respect to mirror using following points.  →   →  VI / m  = VO / m   11  11  →   →  VI / m  = −VO / m   1  1 → →   →  V I / m = VI / m  +V I / m   11  1

Velocity of image with respect to required observer is then calculated using basic equation for relative motion. →

→

→

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Note: If the velocity of the object (w.r.t mirror) is not in a direction normal to the mirror, then the velocity of the object can be resolved into two components one normal to the mirror (vn) and the other along the mirror (vp). The image has velocities –Vn and VP, normal to and along the mirror. Q: 1) Point object is moving with a speed V before an arrangement of two mirrors as shown in figure. Find the velocity of image in mirror M1 w.r.t. image in mirror M2? Solution:

→

→

→

V 1/ 2 = V 1 − V 2 =

2V sin θ

F.B.D

Angle between VI and VI ∴ their magnitude is V. Is 2 θ 1

2

Q: 2) Find the velocity of image of a moving particle in situation as shown in figure. Solution: Analysis:

For component of velocity of image ⊥1/ 2 to mirror →

→

→

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∴ (VI ) ⊥1 / 2 = 2( − 2 ) − 6 = −10 m / s For component of velocity of image parallel to the mirror

(VI ) 11 = 8m / s ∴ Velocity of time VI = (VI )12 + (VI ) 2n = 100 +64 = 164 m

4 ∴ θ = tan −1   5 

Q: 3) Two plane mirror are placed as shown in the figure below:

A point object is approaching the intersection point of mirror with a speed of 100cm/s. The velocity of the image of object formed by M2 w.r.t. velocity of image of object formed by M1 is: Solution:

The components of various velocities are as shown in the figure below

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→

V IM 2 is given by the vector sum of components of velocity of image w.r.t. M2 along the normal and ⊥1 r to the normal.

   V = 1 s 3 i+1 s 3 ic03 j+ 7−1inc0 3o7i+1n0s73 c 3o0 j 7 si 0 o7 ns70 0 s    → ^ ^ ^ ^ 02 0 0 02 0 0 I2 M

 ^ ^ =  − 2 i+ 48 j c8 / sm   →

→

→

V IM 2 , IM1 = V IM 2 − V IM1

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 ^ ^ =  − 1 i+ 24 j 8c /sm e c   Q: 4) In the situation show in figure, find the velocity of image? Solution:

Along x – direction, applying

Vi = Vm = −(V0 −Vm )

[

Vi − ( − 5 cos 30 0 ) = − 10 cos 60 0 − ( − 5 cos 30 0 )

( )

]

^

∴ Vi = − 5 1 + 3 i m / s Along y-direction V0 = Vi ^

∴ Vi = 1 0s i n6 0 = 5 j m / s 0

∴ Velocity of the image

()

^ ^

= − 5 1+ 3 i + 5 j m / s

Q: 5) An object moves with 5m/s towards right while the mirror moves with 1m/s towards the left as shown. Find the velocity of image. Solution: Take → as +ve direction.

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Vi −Vm =Vm −V0

Vi − ( −1) = ( −1) − 5 Vi = −7 m / s ⇒ 7 m / s and direction towards left.

Q: 6) Find the region on y-axis in which reflected rays are present object is at A(2, 0) and MN is a plane mirror, as shown Solution:

A ' = ( 6,0 ) M ' = ( 0,6 ) N ' = ( 0,9 )

Q: 7) An object moves towards a plane mirror with a speed v at an angle 600 to the ⊥1 r to the plane of the mirror. What is the relative velocity between the object and the emage? a) V Solution:

b) →

→

3 V 2

c)

V 2

d)

V 2

→

V OI = V O − V I

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   Vc 6 i− Vs 6o  + 0Vj ci 6 i0+ Vsn 6 oj 0 i 0sn    ^ ^ ^ ^ 0 0 0 0

Q: 8) A ray of light making angle 200 with the horizontal is incident on a plane mirror with itself inclined to the horizontal at angle 100, with normal away from the incident ray. What is the angle made by the reflected ray with the horizontal? Solution: AO = Incident ray OB = Reflected ray

The reflected ray goes along the horizontal. Hence angle made by the reflected ray with the horizontal is zero.

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Q: 9) A ray of light making angle 100 with the horizontal is incident on a plane mirror making angle θ with the horizontal. What should be the value of θ , so that the reflected ray goes vertically upwards? a) 300 b) 400 c) 500 d) 600 Solution:

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Number of Images Formed by two Inclined-Plane Mirrors: a) When mirror are parallel: In this case, infinite images are formed due to multiple reflections.

b) When mirror are perpendicular: In this case, three images are formed. The ray diagram is shown.

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Note that the third image is formed due to rays undergoing two successive reflection. Also, object and its images lie on a circle whose equation is given by x 2 + y 2 = a 2 + b 2 . When an object is placed in front of arrangement of three mutually perpendicular mirror, then total seven images are formed. Further, object and its image lie on a sphere whose equation is given by x 2 + y 2 + z 2 = a 2 + b 2 + c 2 , where a, b and c are co-ordinates of object.

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Minimum size of Mirror to see Full-Image:

AB is the person with E as his eyes, M1 M2 = plane mirror infront of him. For the length of the mirror to be minimum, the rays coming from the extreme top and bottom portions of his body. (i.e.) A and B, Should after reflection, be able to just enter his eyes. The light ray AM, is incident ray and M1E the reflected ray. ∠AM 1 N1 = ∠EM 1 N1 So As ∆' s AM 1 N1 and EM 1 N1 are similar. ∴ M 1 E1 = x

Say = N1 E =

1 ( AE ) 2

……..(1)

Similarly the light rays BM2 and M2E are incident and reflection rays respectively So ∠BM 2 N 2 = ∠EM 2 N 2 S ∴ ∆ s BM 2 N 2 and EM 2 N 2 are similar ∴

M 2 E1 = y ( Say ) = N 2 E =

1 ( BE ) 2

………(2)

Adding equation (1) and (2) yield x + y = length of mirror = 1 ( AE + BC ) = 1 ( AB ) = 1 (Height of person) 2 2 2

Note:- Minimum size is independent of distance between man and mirror. Q: 1) A plane mirror is inclined at an angle θ with the horizontal surface. A particle is projected from point P (see fig.) at t = 0 with a velocity v at angle α with the horizontal. The image of the particle is observed from the frame of the particle projected. Assuming the particle does not collide the mirror, find the (a) time when the image will come momentarily at rest w.r.t. the particle (b) path of the image as seen by the particle. Solution: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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(a) The image will appear to be at rest w.r.t. the particle at the instant, the velocity of the particle is parallel to the mirror.

Vy

= tan θ Vx V sin α − gt = tan θ V cos α V cos α ( tan α − tan θ ) t= g

(b)

St. line ⊥1 r to mirror

Q: 2) An a oblong object PQ of height ‘h’ stands erect on a flat horizontal mirror. Sun rays fall on the object at a certain angle. Find the length of the shadow on screen placed beyond the shadow on the mirror.

Solution:

PS = Shadow on the mirror P’ Q’ = Inversed shadow of PQ on the screen

α = angle of incidence Let Then PS = h tan α and QS = h sec α From the properly of image P’ Q’ = 2( h sec α) cos α = 2h

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Q: 3) A plane mirror is placed at parallel of y-axis, facing the positive x-axis. An object starts form

   2 i+ 2 j  m / s   ^ ^

(2m, 0, 0) with a velocity of

. The relative velocity of image with respect to object is

along Solution:

→

→

V0 = VI = →

V0 =2

(2)2

+( 2 )

2

2 m/s

Relative velocity of image with respect to object is in negative x-direction as shown in figure.

Q: 4) A reflection surface is represented by the equation x 2 + y 2 = a 2 . A ray traveling in negative x-direction is directed towards positive y-direction after reflection from the surface at some point ‘P’. Then co-ordinates of point ‘P’ are Solution: From figure

9 9 , y= 2 2 q   q ∴ P = ,  2  2 x=

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Q: 5) A ray is traveling along x-axis in negative x-direction. A plane mirror is placed at origin facing the ray. What should be the angle of plane mirror with the x-axis so that the ray of light offer reflecting from the plane mirror passes through point (1m, 3 m )? Solution:

Q: 6) Two plane mirror A and B are aligned parallel to each other as shown in the figure. A light ray is incident at an angle 300 at a point just inside one end of A. The plane of incidence coincides with the plane of the figure. The maximum number of times the ray undergoes reflection (including the first one) before it emerges out is____ Solution: d = 0.2 tan 30 0 =

0.2 3

∴ Max . No . of reflection

=

2 3 = 30 0.2 / 3

REFLECTIN FROM CURVED SURFACE: (Spherical – Surface only) A curved mirror is a smooth reflecting part of any geometry. The nomenclature of curved mirror depends on the geometry of reflecting surface. There are different types of curved mirror like paraboloidal, ellipsoidal, cylindrical, spherical ….etc.

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•

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Sign-Conversion:

Rules for Ray-Diagrams 1) A ray of light parallel to principal axis passes (or) appears to pass through four after reflection.

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2) A ray of light passing through focus (or) appears to pass through focus becomes parallel to principal-axis after reflection.

3) A ray of light passing through (or) appears to pass through centre of curvature is reflected back.

4) A ray of light hitting pole is reflected making equal angle with principal oxis

Note: 1) Focal length and radius of curvature of plane mirror = 2) Concave mirror = Convergent mirror Convex mirror = Divergent mirror Relation between focal-length and radius of curvature: •

•

α

R   f = 2  Both for concave and convex mirror.

Mirror formula (or) Mirror Equation: The relation between u, v and f of a mirror is known as mirror formula  1 1 1  f = u +v  

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Relation between the speeds of object and image formed by a spherical mirror We know that, mirror formula is given by Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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1 1 1 + = u v f

……….(1)

Differentiating both sides w.r.t. time (t), we get 1 du 1 dv − 2. − . =0 u dt v 2 dt 1 dv 1 dv . =− 2 . =0 2 v dt u dt dv v 2 dy …………(2) =− 2 . dt u dt dv = vi = speed of time dt du = v0 = speed of object dt 2 v  ∴vi =   .v0 ………….(3) u 

Since

From equation (1), v=

•

uf u− f

v f = u x− f

Hence equation (2) become  f   .v0 vi = − u− f  Linear magnification: It is defined as the ratio of the size (or) height of the image to the size (or) height of the object. m=

•

( or )

size of image height = size of image height

of image of image

I  ∴ M =  o 

Magnification produced by concave mirror:

A ' B ' =image of object AB ∆S ABp and A' B ' p are similar

∴

A'B ' PA ' = AB PA

…………(1)

Applying sign conversion AB = +0 A' B ' = −1

PA = −u PA ' = −v

∴Equation (1) can be rewritten as Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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−I −v = + O −u I v =− O u v  ∴ m = −  same for convex-mirror also. u 

•

Magnification in terms of u, v and f 1

1

1

a) As we know that u + v = f Multiplying both sides by ‘u’ we get u u u + = u v f u u 1+ = v f u u u−f = −1 = v f f v f ∴ = m u−f

Since m = − ∴m =−

v u

f u−f

( or )

1

1

 f  m = f −u    1

b) As we know that u + v = f Multiplying both sides by V, we get v v v + = u v f v v +1 = u f v v v−f = −1 = u f f

Since m = −

v u

v − f = −  f 

 f −v   ( or ) m = f 

Note: a) +ve magnification mean both object and image are upright b) –ve magnification means, object and image have different orientation (i.e.) if object is upright, then image is inverted.

•

LATERAL-MAGNIFICATION (mL) length of image Li mL = = length of object L0 Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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For extended objects the lateral magnification can be obtained by independently imaging the two end points and calculating the length of the image. There is no direct formula to obtain the magnification. However, if the length of the object is small, them the lateral magnification can be directly obtained from equation 1 1 1 + = u v f

Differentiating both sides, we get du dv − 2 − 2 =0 u v  dv  v2  du = − u 2 = mL   

Q: 1) What do we do if the size of the object is large as compared to the distance u? Analysis: For extended object

V A − VB u A − uB For tip A m2 =

u = −( x + L )

f =−

R 2

V =V B 1 1 1 + = v u f

1 1 −2 − = from which VB can be obtained vB x + l R

∴Subtracting VB from VA, we can calculate the length of the image.

•

Combinations of mirrors: What do we do if we have a combination of mirror? If an object is placed between the mirrors, how do we find the final position of he image? Analysis: In such situations, we need to simply solve for the reflection at each of the mirror keeping in mind that the image formed by the first mirror is the object of the second mirror and so on. Case must be taken to correctly apply the sign conversion at each of the mirror. Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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Q: 1) Find the velocity of image in situation as shown in figure?

  V0 =  9i+ 2i = 1 i m/1s   → ^ ^ ^

Solution:

→

^

Vm = − 2i m/s ∴m =

f − 20 = = −2 f − u − 20 − ( − 30 )

∴(VI / M

)11

→  =−m 2 V O / M   11

= -(-2)2 11 = -44

^

i

^

i m/s.

−  −  ∴ (VI / m ) = V I / m  +V I / m   n  I

= −( − 2 ) 12 x = −24 j m / s

  −  −  ∴V I / m  = V I / m  + V I / m   n  n  I −

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 ^ ^ =  − 4 i− 24 j m4 / s   −  →  ^ → ^ VI =  V I/m  = V m =  − 4 i− 24 j  =42i      ^ → =  − 4 i− 26 j  4   Q: 2) A thin rod of length

f is placed along the principal axis of a concave mirror of focal-length ‘f’ 3

such that its image just touches the rod, calculate magnification? Solution: Since image touches the rod, the rod must be placed with one end at centre of curvature. Case –I Case –II

f  −5f  u = − 2 f −  = 3 3  f = −f

 5f − ( − f ) u f 5f 3   v= = = − 5 f u− f 2 − (− f ) 3

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5f − − (− 2 f ) V A − VC 3f 3 ∴ m= = =− −5f u A − uC 2 − (− 2 f ) 3 f  7f  x = − 2 f +  = − 3 3  f = −f

−7 f   (− f ) uf 7f 3   V= = =− −7 f u− f 4 −(− f ) 3 −7 f −( − 2 f ) V A − VC 3 ∴M = = 4 =− −7 f u A − uC 4 −(− 2 f ) 3

CONCEPTUAL POINTS It a hole is formed at the center of mirror, the image position and size will not change. • The intensity will reduce depending on the size of the hole. For all object positions a convex-mirror forms a virtual and erect image • PROBLEMS OF MIRRORS Q: 1) A short linear object of length ‘b’ lies along the axis of a concave mirror of focal-length f, at a distance u from the mirror. The size of image approximately is 2

Solution:

 f  V   M axial =   =  u   f −u   f  I  =  O   f −u  I  f   = b  f − u 

2

2

2

2   f     ⇒ I = b   f − u  

Q: 2) Two spherical mirrors M1 and M2 one convex and other concave having same radius of curvature R are arranged coaxially at a distance 2R (consider their pole separation to be 2R). A bead of radius a is placed at the pole of the convex mirror as shown. The ratio of the sizes of the first three images of the bead is Solution: The first image is formed due to the reflection from concave mirror M2

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1 1 2 + = V1 ( − 2 R ) − R 1 1 4 = − V1 2R 2R 1 −3 = V1 2R 1 −2 R V1 3

 −2 R  ⇒m/ = − 3  2R  

   = −1 3   

.

⇒ object distance = 2 R − 2 R = 4 R 3 3 1 1 2 + = V2  4 R   R  −      3  2 1 2 2 = + V2 R 4R 4R V2 = 11 4R − V 3 m2 = − 2 = 11 = − 4 R u2 11 3 3 ∴m2 = 11 3 a a So radius of second image ⇒a 2 = . = 11 3 11 a Similarly radius of third image is a3 = 41 1 1 1 ∴ : : Answer 3 11 41

Q: 3) When an object is placed at a distance of 60cm from a convex spherical mirror, the magnification produced is 1/2. where should the object be placed to get a magnification of 1/3? u = −60 cm Solution: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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V m =− u 1 V =− 2 − 60

(or) V = +30 cm

1 1 1 1 1 1 = + =− + = f u v 60 30 60 ∴ f =+60 cm ∴

In second case m=

As

1 V =− 3 u

100 V = −

u 3

1 1 1 + = u v f

1 3 1 − = u u 60

u = −120 cm

Q: 4) Two objects A and B when placed one after another in front of a concave mirror of focal-length 10cm, form images if same size. Size of object A is 4 times that of B. If object A is placed at a distance of 50cm from the mirror, what should be the distance of B from the mirror? Solution: For object A For object B h h' f f m= 2 = m' = 2 = h1 f − u1 h '1 f − u1

m h2 h11 f − u2 ∴ = × 2 = m ' h1 h2 f − u1 As

h1 = 4h11 and h2 = h21 , f = −10 cm u1 = −50 cm

∴

1 − 10 − u 2 = 4 − 10 + 50

u 2 = −20 cm

Q: 5) A concave mirror of focal length 10cm is placed at a distance of 35cm form a wal. How far from the wall should an object be placed to get in image on the wall? Solution:

f = −10 cm ,

V = −35 cm

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∴

1 1 1 = − u f V

=−

1 1 1 + =− 10 35 14

∴ u =−14 cm

Distance of the object form wall = 35 – 14 = 21 cm Q: 6) An object is placed at a distance of 36cm form 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 if 24cm from the object, find the radius of curvature of the convex mirror. OP = u = −36 cm Solution: V = PI = +12 cm

1 1 1 1 1 −1 + 3 1 = + = + = = f u V 36 12 36 18 ∴ f =18 cm ∴ R = 2 f = 2 ×18 = 36 cm ∴

Q: 7) A convex mirror of focal length ‘f’ forms an image which is the object which is Solution:

1

1 times the object. The distance of n

1 times the object. The distance of the object from the mirror is n

V

η =+ =− η u V =−

u

η

1 1 1 ∴ = + f V u 1 1 1 = + f η  −u     η  Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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u = −(η −1) f

Q: 8) An object of size 7.5cm is placed in front of a convex mirror of radius of curvature 25cm at a distance of 40cm. The size of the image should be Solution:

η=

I f = O f −u

u = −40

( R / 2) = ( 25 / 2) I = 7.5 ( R / 2) − u ( 25 / 2 ) − ( − 40 )

I =+1.78 cm

Q: 9) The image formed by a convex mirror of focal length 30cm is a quarter of the size of the object. The distance of the object from the mirror is Solution:

m=

f f −u

+ 30  1 +  =  4  + 30 − u u = −90 cm

Q: 10) A concave mirror of focal length f (in air) is immersed in water ( µ = 4 / 3) . The focal length of the mirror in water will be a) f Solution:

b)

4 f 3

c)

3 f 4

d)

7 f 3

On immersing a mirror in water, focal length of the mirror remains uncharged.

Q: 11) An object is 20cm away form a concave mirror with focal-length 15cm. If the object moves with a speed of 5m/s along the axis, then the speed of the image will be Solution:

1 1 1 − = V 20 −15

V =−60 cm

2

V  Vi = −  . V0 u  2

 60  =  .( 5)  20  = 45 m / s

Q: 12) A concave mirror is placed at the bottom of an empty tank with face upwards and axis vertical when Sun-light falls normally in the mirror, it is focused at distance of 32cm form the mirror. If  

4 3

the tank filled with water  µ =  Ans:

upto a height of 20cm, then the Sunlight will now get

focused at 9cm above water level

Q: 13) A small piece of wire bent into an ‘L’ shape with upright and horizontal portions of equallengths, is placed with the horizontal portion along the axis of the concave mirror whose radius Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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of curvature is 10cm. If the bend is 20cm from the pole of the mirror, then the ratio of the lengths of the images of the upright and horizontal portions of the wire is Solution:

f =

R 10 = = 5 cm 2 2

For part PQ  f  L 1=   f −u   L0  

  L −5  × L0 = − 0 =  3  − 5 − ( − 20 ) 

For part QR 2

 f  L 2=  f −u   L0   2

  L0 −5 =  − 5 − ( − 20 )   × L0 = 9  

L1 3 = L2 1 CONCEPT OF NEWTON’S FORMULA (FOR A MIRROR) In this formula, the object and image distance are expressed w.r.t. focus. Consider an object O kept beyond ‘C’ of a concave mirror, and whose image is formed at I with in C. ∴

•

Let OF = x and IF = y From triangle OMC OC OM OM = = sin θ sin ( π − α ) sin α

…………(1)

And from triangle ICM IC IM = sin θ sin α

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Dividing equation (1) and (2) yields OC OM OP = = (since M is close to P) IC IM IP

(or)

x−f x+ f = f −y f +y 2 x f − f +x y − f y = x f −x y + f (or) x y = f 2

∴f =

1)

As x y = f

2

2

−f y

x y

1

(or) xα y

(i.e.) The distance of object and image form the focus are inversely proportional to each other. In other words, the more the object distance (from the focus), the less will be the image distance (from the focus) and vice versa 2)

If x → 0; y → ∞ and if x → ∞; y → 0 . If the object is at focus the image is a far off distance and vice-versa.

3)

From xy = f 2 ; if x = f , then y = f . Thus, if the object be at C, then image will also be at C (for a concave mirror) and if the object is at P, then the image will also be at P (for a convex mirror)

4)

Since f 2 is necessarily +ve for both types of mirror, so x and y bear the same sign, which implies that both the object and the image always lie an the same side of focus.

A) GRAPH OF |x| Versus |y| : Since xy = f 2 represents a rectangular hyperbola, existing in the first and third quadrant ( f 2 being positive). ∴The graph of |y| vs |x| will be a rectangular hyperbola existing only in the first quadrant.

B) GRAPH OF U Versus V : Since xy = f 2 ∴( u − f

)( v − f ) = f 2

For a convex mirror, u is always negative and V is always positive. Further ‘f’ is also positive. and V = y we have

∴Putting u = x (x − f ) ( y − f ) =

f

2

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This is the equation of a rectangular hyperbola with its origin shifted to ( f , f ) and ’x’ being always negative while y lies between O and f. (see figure) for a concave mirror, u is always negative, v can be positive (or) negative, ‘f’ is negative

∴ u = x, v = y

and

f =−f

We have, form

(u − f ) ( v − f ) = f 2 (u + f ) ( y + f ) = f 2 [ x − ( − f )] [ y − ( − f )] = Or

f

2

Evidently it is again an equation of a rectangular hyperbola with origin of coordinates shifted to the point (-f, -f) (see figure ↓)

3) GRAPH OF

1 1 VERSUS v u

From mirror formula 1 1 1 + = v u f

Putting

1 1 = x and = y , we have u v

x+y =

1 y

It is the equation of a straight line having a slope +1 (or) -1 according as u and v bear the same 1

(or) opposite signs. The intercepts on x and y axis are each + f object and image are to the right (or) left of the mirror.

( or )

−

1 according as the f

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For a concave mirror: u is always –ve v can be positive (or) negative and f is –ve.

For a convex mirror u is always negative v is always positive and f is always positive.

•

CONCEPT OF CRITICLA ANGLE When a ray of light is traveling form denser medium to rarer medium, it get refracted and the ray derivates away form the normal. If we keep increasing angle of incidence then at an angle, the angle of refraction becomes 900 . This is known as Critical –Angle (c). When angle of incidence is increased, further the ray gets reflected back in the same medium. This phenomena is known as T.I.R.

According to Snell’s Law b

µa =

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⇒ ⇒ ⇒

sin iC sin 90 0

b

µa =

b

µa = sin iC

b 1   µa =  sin iC  

C depends on colour and and temp ∴CRed < Cviolet CRed > Cviolet If temp ↑C ↓ Q: 1) The sum (diameter d) subtends an angle θ radius at the pole of a concave mirror of focal length f. Find te diameter of the image of sun formed by mirror? Solution:

1 1 1 + = v u f

we get

1 1 1 = (∴u is very large so ≈ 0 v −f u Or v = − f

It means image is formed at focus Taking ' f ' as radius and using θ=

 when  = d and r = f r

∴θ =

d or d =θ f f

• REFRACTION AT SPHERICAL SURFACE:

le s OBC and IBC From ∆

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We have i = α + β And β = r + r ( or ) r = β − r From Snell’s law µ2 sin i = µ1 sin r µ1 sin i = µ2 sin r For small angle of incidence I, we can write sin i ≈i and

sin r ≈r

∴ µ1i = µ2 r µ1 [α + β ] = µ 2 [ β − r ] As ‘i’ is small, and so α, β and r are also small. Thus (α + β ) = tan α + tan β h h = + −u + R h h And ( β − r ) = − R V h   h h h ∴ µ1  + = µ2  −   − u + R  R v 

After simplifying we get µ 2 µ1 µ 2 − µ1 − = v u R µ2 µ2 −1 µ1 1 µ1 − = v µ R 1

Q:

µ2 1 1 µ2 − 1 − = → This formula is derived for convex surface and for real v u R

Image From denser to rarer medium µ1 µ 2 µ1 − µ 2 − = v u R How can we derive a mathematical expression for the equation of a ray in the medium? The medium is of variable refractive index. Ray of light is incident at an angle α at air medium interface? Analysis: Here two cases a rise. Refractive index is varying either as function of y (or) function of x. Case-I: µ = f ( y ) (i.e.) Refractive index varies with y At some height h angle of incidence is θy and refractive index is f ( y ) from Snell’s Law

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µ sin θ = constant ∴ 1 ×sin α = f ( y ) sin θy

……..(1)

Slope of curve at A dy = tan ( 90 − θy ) dx dy ⇒cot θy = dx

From equation (1) dy = dx

{ f ( y)

− sin 2 α sin α 2

}

Case-I: µ = f ( y ) (i.e.) Refractive index varies as function of x. According to Snell’s Law

µ sin θ = constant

For initial refraction at the air medium interface 1 × sin α = µ0 sin ( 90 − θ 0 ) sin α = µ0 sin ( 90 − θ 0 ) ∴sin α = µ0 cos θ0 Here θ0 angle of refracting ray at point A with OX sin α cos θ0 = So µ0

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And

sin θ 0 =

1 − sin 2 α µ02

Now Snell’s Law at M gives f ( x ) sin θx = µ0 sin θ0 Or

sin θ x =

µ0 sin 2 α 1− f ( x) µ02

µ02 − sin 2 α f ( x) Now slope of tangent at M is given as ∴ sin θ x =

dy = tan θx dx

dy = dx

µ 02 − sin2 α { f ( x )} 2 − µ 02 + sin2 α

Q: 1) If µ = 1 + y and ray of light is incident at grazing incidence at origin, then find equation of path of refracted ray. Solution: We can use result derived above for which f ( y ) = 1 + y and α = 90 0 dy ∴ =y dx x2 y= 4

So

1/ 2

Q: 2) An object is at a distance 25cm form the curved surface of a glass hemisphere of radius 10cm. Find the position of the image and draw the ray diagram ( µg = 1.5) Solution: For refraction at first face

µ 2 µ1 µ 2 − µ1 − = v u R

µ = −25 cm µ1 = 1

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µ2 = 3 / 2 R = 140cm

∴V =150 cm

∴The rays are converging beyond of at 140cm form Q. Again refraction takes place at the plane surface. For refraction at second face R = ∞, µ2 = 1.5 µ1 = 1 µ = +140 cm

v =?

Using

µ1 µ 2 µ1 − µ 2 − = v u R

1 1.5 1 −1.5 − = v +140 ∞ v = 93 .3

∴The ray meet axis at 93.3cm form point Q.

•

PROBLEMS ON REFRACTION A light ray is incident at an angle of incidence double that of refraction on one face of a parallel sided transparent slab of refractive index ' µ' and thickness ‘t’. Find the lateral displacement of the ray? i =2 r Solution: 1)

t sin ( i − r ) t sin ( 2r − r ) sin r = =t = t tan r cos r cos r cos r sin i µ= sin r sin 2r 2 sin r. cos r µ= = sin r sin r µ = 2 cos r D=

As

cos µ =

µ 2

2

2 ∴ tan r =  µ  −1   2

2 ∴ D =t  µ  −1   ∴ D=

t 4 − µ2

µ

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Q: 2) A light ray is incident on a transparent slab of R. I. µ = 2 , at an angle of incidence π / 4 . Find the ratio of the lateral displacement suffered by the light ray to the maximum value which it could have suffered? Solution:

i=

π , 4

µ= 2

sin i sin π / 4 =µ ⇒ = 2 sin r sin r

∴r = π / 6

π π  t sin  −  t sin ( i − r ) 4 6 D= = π cos r cos 6 π π π 

= t  sin 4   1 =t  −  2 t D= 6

(

− cos

4

. tan

 6

1 1  ×  2 3

)

3 −1

∴

(

D 3 2− 6 = t 6

)

Q: 3) Light of a certain colour has 2000 waves in one millimeter of air. Find the number of waves of that light in one millimeter length of water and glass respectively? λa = wavelength in air Solution: λm = wavelength in medium The number of waves of that light in a length of ‘d’ will be n1 =

And

n2 =

d

λa d

λm

n2 λa = = µm n1 λm ∴ n2 = µm × n1 ∴

n2 = 1.33 × 2000 = 2660 in water n2 = 1.50 × 2000 = 3000 in glass

Q: 4) Light from a sodium-lamp ( λ = 58 nm ) traverses a distance of 60m in a chloroform ( µ =1.45 ) in a certain time. Determine the time difference occurring when light happens to traverse the same length in ethyl ether ( µ =1.35 ) d 60 m (1.45 ) t= tchloroform 1 = Solution: cm 3 ×10 8 m / s For ethyl ether 60 m(1.35 ) t e.r2 = 3 × 10 8 m / s ∆t = t1 − t 2

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60 m (1.45 −1.30 ) 30 ×10 8 = 2.0 ×10 −8 sec

=

Q: 5) A rectangular glass slab of thickness 12cm is placed over a small coin kept on a table. A thin transparent beaker filled with wager to a height 6cm & placed over the block. Find the apparent shift of the position of the coin, when viewed from a point directly above it?

Solution:

 1  S1 = t1  1 − µ   1  

2  = 12 1 −  3  =4 cm

 1 S 2 = t 2 1 − µ2  3  = 61 −  4 

  

= 1.5 cm

∴ S = S1 + S 2 = 4 +1.5 = 5.5 cm

Q: 6) The time taken by light to cover a distance of 9mm in water is____ Solution:

3 ×10 8 9 = ×10 8 m / s 4/3 4 d 9 ×10 −3 × 4 t= = = 4 ×10 −11 sec Cw 9 ×10 8 Cw =

= 4 ×10 −11 ×10 9 ns =0.04 ns

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Q: 7) A ray incident at an angle of incident 600 enters a glass – sphere of R.I µ = 3 . The ray is reflected and reflected and refracted at the farther surface of the sphere. The angle between reflected and refracted rays at this surface is_____ Solution:

µ=

sin i sin r

3 sin 60 1 sin r = = 2 = µ 3 2 0

∴r =30 0

PC = QC ∴∠CPQ = ∠PQC = ∠r = 30 0

Angle between reflected ray QR and refracted ray QS at the other face = 180 − r − 60 0 = 180 − 30 0 − 60 0 = 90 0

REFRACTION AT SPERICAL SURFACES • Q: 1) Sunshine recorder globe of 30cm diameter is made of glass of refractive index µ =1.5 . A ray enters the globe parallel to the axis. Find the position from the centre of the sphere where the ray crosses the principal axis? Solution:

For first refraction (Rares to deuces) u = −∞ µ2 = 1.5 µ1 = 1 R = +15cm µ µ µ − µ1 ∴ 2− 1 = 2 v 2 R

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µ 2  µ 2 − µ1  µ 1 = + V  R  u

1.5  1.5 −1  1 1 = = + V 15 ( − ∞ ) 30  

V = 45 cm

For second refraction (douses to rarer) R = −15 cm ,

u ' = ( 45 −30 ) =15 cm

µ1 µ2 µ1 − µ2 Using V ' − u ' = R

1 1.5 1 −1.5 1 − = = V ' 15 −15 30

( or ) V ' = 30 4

=7.5 cm

∴Distance of image from centre of globe is (15 + 7.5) = 22.5cm Q: 2) A particle executes a simple harmonic motion of amplitude 1.0cm along the principal axis of a convex lens of focal length 12cm. The mean position of oscillation is at 20cm from the lens. Find the amplitude of oscillation of the image of the particle? Solution: When the particle is

At R

u = −19 cm V1 = ? f =12 cm

1 1 1 − = V1 u f 1 1 1 = − V1 12 19 12 ×19 V1 = 7

1 1 1  = +  f u V 7 = 12 ×19

When the particle is at left extreme position u = −21 cm, V2 = ? 1 1 1 ( or ) − = V2 u f

f = 12 cm 1 1 1 1 1 9 = + = − = V2 f u 12 21 12 × 21

∴Amplitude of oscillation of image

=

V1 −V2 2

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=

1 12 ×19 12 × 2  − cm 2  7 9 

= 2.2857

cm

Q: 3) A point object is moving with velocity 0.01m/s on principal axis towards a convex lens of locallength 0.3m when object is a distance of 0.4m form the lens, find a) Rate of charge of position of the image and b) Rate of charge of lateral magnification of image Solution: Differentiating the

1

1

1

a) Equation f = v − u w.r.t. time 1 dv 1 du 0=− 2. + 2. V dt u dt dv V 2 du = 2. dt u dt 1 1 1 = − ⇒ V = 120 cm 30 V − 40 du u = −40 cm , = 0.01 m / s dt dv 120 ×120 ∴ = × 0.01 m / s dt 40 × 40 = 0.09 m / s

2

dv V 2  V = 2 = 1 −  b) M = du u f    dm Vd  V = 2 1 − f   dt  1 − f dt   

   

 V  1 dv = −2 1 − f   f . dt   2 V  dv 2  120  −1 =−  1−  =− 1 − 0.09 s   f  f  dt 30  30  = 0.018 / sec

Q: 4) Find the position of the image formed by the lens combination given in figure?

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Solution:

Image formed by first lens 1 1 1 − = V1 u1 f1 1 1 1 + = V1 30 10

1 1 1 − = V1 − 30 10

( or )

( or ) V1 =15 cm

The image formed by the first lens server as the object for the second. This is at a distance of (15 – 5)cm i.e. 10cm to the right of the second lens. It is a virtual object Now

1 1 1 − = V2 10 −10 1 1 1 =− + =0 V2 10 10

V2 = ∞

The virtual image is formed at an infinite distance to the left of the second lens. This acts as an object for the third lens. 1 1 1 − = V3 − ∞ 30 1 1 = V3 30 V3 = 30 cm

The final image is formed 30cm to the fight of the third lens. Q: 5) Two Plano-concave lenses of glass of refractive index 1.5 have radii of curvature 20cm and 30cm respectively. They are placed in contact with the curved surfaces towards each other and the space between them is filled with a liquid of refractive index 5/2. Find the focal length of the combination. Solution: For first plano concave lens

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− R2 − 20 − 20 = = = −40 cm µ1 − 1 1.5 − 1 0.5 For second plano concave lens − R2 − 30 − 30 f2 = = = = −60 cm µ1 −1 1.5 −1 0.5 The focal length of the liquid lens is given by f1 =

 1 1 1   = ( µ2 −1) +   f3  R1 R2 

R1 = 20cm,

R2 = 30cm

µ2 = 5 / 2

∴ f 3 =8 cm 1 1 1 1 1 1 1 1 ∴ = + + =− − + = cm f f1 f2 f3 40 60 8 12 ∴f =12 cm

Q: 6) Given the object image and principal axis find the positions and nature of the lens Solution: First join the object and image

If the one point is above the optical axis and the other below it, then the lens is always a convex lens.

If object and image points are Above the principal axis and image point is higher, then the lens is convex and is present between the image and object points. Other wise the lens is concave. Q: 7) For the given positions of the objects and the image in figure determine the location and the nature of the lens used? Solution:

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Q: 8) A ray of light passes through a medium whose refractive index varies with distance as x  µ = µ0 1 +  . If the ray enters the medium parallel to the x-axis, what will be the trajectory of  a the ray and what will be the time taken for the ray to travel a distance a? Solution: The ray enters normally and proceeds along a straight line. At a distance ‘x’ in the medium consider a slab of thickness “dx”. Velocity of the light ray at this point is

Time taken to cross the distance ‘dx’ is dx dx dx x  C dt = = = × µ 0 1 +  x C V   a µ 1 +   a ∴ Total time of travel is a

t =∫ 0

dx  x µ0 1 +  C a 

3 µ0 a = 2 C

Q: 9) A fish is rising up vertically inside a pond with velocity 4cm/s and notices a bird which is diving vertically downward and in velocity appears to be 16cm/s (to the fish). What is the velocity of the diving bird, if R. I of water is 4/3? Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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Solution:

Vb f = Vb − ( −V f

) ∴µ =

16 =Vb +V f

Vb V

4 12 = 3 V 3 V = ×12 = 9 cm / s 4

16 = Vb + 4 Vb =12

Q: 10) Solar rays are incident at 450 on the surface of water ( µ = 4 / 3) . What is the length of the shadow of a pole of length 1.2m erected at the bottom of the pond, if the pole is vertical assuming that 0.2m of the pole is above the water surface? Solution: Applying Snell’s law at point c

1 × sin 45 0 = sin θ =

4 sin θ 3

3 4 2

Here AE = CD = 0.2m tan θ = sin θ = 3 4 2

=

BC BE = = BE CE 1 BE 1 + ( BE ) BE

2

1 + ( BE )

2

BE = 0.625 m

∴ The length of shadow = AB = AE + EB = 0.2+0.625 = 0.825 Q: 11) In a lake, a fish rising vertically to the surface of water uniformly at the rate of 3m/s, observes a bird diving vertically towards the water at a rate of 9m/s vertically above it. The actual velocity of the dive of the bird is_______ ( µ = 4 / 3) Solution: Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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µ=

A.D R.D

y' y ∴y ' =µy ∴h =x +y ' h = x +µ y

µ=

Differentiating dh dx dy = +µ dt dt dt dy 9 =3 +µ dt dy 6 = = 4.5 m / s ( 4 / 3) dt

Q: 12) A convex lens of focal length 0.2m is cut into two halves each of which is displaced by 0.0005m and a point object is placed at a distance of 0.3m form the lens, as shown in figure. The position of the image is ________ Solution:

1 1 1 = − f v u 1 1 1 = + v f u

u = −0.3m 1 1 1 = − v 0.2 0.3

f = 0 .2

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v = 0 .6 m

Q: 13) A pole 5m high is situated on a horizontal surface. Sun rays are incident at an angle 30 0 with the vertical. The size of shadow in horizontal surface is______ Solution:

tan 30 0 =

BC 5

BC = 5 tan 30 0 =

5 m 3

Q: 14) The Sun subtends an angle α = 0.50 at the pole of a concave mirror. The radius is curvature of concave mirror is R = 1.5m. The size of image formed by the concave mirror is_____ Solution: As Sun is at infinity image is formed at the focus of mirror

1)

1 ×0.5 0 ×108 2 1 π = × 0.5 × ×105 2 180 =

= 0.654 cm ∠POQ = α =

DS Or u

Di = magnification = v = f DS u u Di R = DS 2u Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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Di =

•

R ×DS 2u

MAGNIFICAIOTN IN CASE OF CURVED SURFACE: Consider an extended object “QO” placed ⊥1 r to the principal axis at the point O. The image of the point ‘O’ is formed at I. Image of extended object is ‘IQ’ ∴ From ∆les POQ and PIQ ' we can say that I Q' OQ and tan θ2 = P I PO But since θ1 and θ2 are very small, we can approximate Also µ1 sin θ1 = µ2 sin θ2 Sign conversion h0 = +OQ µ1θ1 = µ2θ2 OQ IQ ' hi = −IQ ' ∴µ1 . =µ2 . PO PI

tan θ1 =

h0 ( − h i) = u2 . (− u) (v) h i µ1 V = h0 µ2 u

∴ µ1

•

u = −PO

V = + PI

LATERAL MAGNIFICATION

The lateral magnification is mL =

dv dx

Which can be obtained by differentiating equation µ 2 µ1 ( µ 2 − µ1 ) − = v u R Thus −

µ 2 dv µ1 . + =0 v 2 du u 2

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•

REFRACTION AT PRISM A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. a) Calculation of angle of Derivation:

In passing through the prism, ray KL suffers two refractions and has tweed through an LQPN = δ . (with is angle of direction) In ∆ PLM , δ = ∠PLM + ∠PML δ = ( i1 − r1 ) + ( i2 − r2 ) δ = ( i1 + i2 ) − ( r1 + r2 ) ……………(1) In ∆ OLM ∠O + r1 + r2 = 180 0 ……………(2) In Quadrilateral ALOM, As (∴Sum of 4 angles of a Quad = 360) ∠L + ∠M = 180 0 ∴ A + ∠O =180

0

∴using equation (2) we can write

∠O + r1 + r2 A + ∠O r1 + r2 = A …………….(3)

Putting (3) in (1) δ = ( i1 + i2 ) − A ……………(4) (for bigger refracting angles) If ' µ' is the refractive index of the material of the prism, then according to Snell’s Law sin i1 i1 µ= = (when angles are small) sin r1 r1 ∴ i1 − µ r1 similarly i2 = µ r2 Putting the above two in equation (4) we get δ = ( u r1 + u r2 ) − A δ = µ ( r1 + r2 ) − A ∴δ = ( µ −1) A (when refracting angle are small) This is the angle through which a ray derivate on passing through a thin prism of small refracting angle A. Prism formula: In minimum derivation position i1 = i2 = i and r1 = r2 = r ∴ from equation (3) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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r +r = A 2r =A

And from equation (4) δm = (i + i) − A δm = 2 i − A

∴ A + δm = 2i

( or )

i=

∴ from Snell’s Law A +δm sin sin i 2 µ= = sin r sin ( A / 2 )

A + δm 2

Note : The deviation through a prism is maximum when i1 = 900. Thus δm = 90 + i2 − A

•

MAXIMUM DEVIATION:

For

i1 = 90 0 ∴ i2 = ?

At surface BD µ1 sin i1 = µ.sin r1 1×sin 90 0 = µ.sin r1 1  ∴sin r1 =  µ    1 r1 = sin −1  µ  or  

( r1 = θC )

We know that r1 + r2 = A r2 = ( A − r1 ) r2 = ( A − θ C ) Flat No.301, Rukkus & Yellus Arcade, Fever Hospital Road, Barkathpura, Hyd-500065. Ph: 040 – 64606657, 9391326657.

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For surface CD µsin r2 = sin i2 sin i2 = µ sin ( A −θC )

i2 = sin −1 [ µ sin ( A − θ ) ] CONCEPT: Sometimes a part of a prism is given and the keep on thinking whether how should we proceed? To solve such problems first complete the prism then solve as the problems of prism are solved

Defects in Image formed by leuses: The defect in the lens on account of which it does not form a white point image of white point object is defined as Aberration.

Axial Chromatic Aberration: The variation of the image distance form the lens with the colour measures axial chromatic Aberration. Lateral Chromatic Aberration: The variation in the size of the image with colour measures the lateral transverse chromatic aberration. Concept of Dispersion of Light: Dispersion of light is the phenomenon of splitting of a beam of white light into its constituent colours on passing through a prism. (accuse due to wavelength). The band of seven colours so obtained is called the visible spectrum. The order of colours from the lower end of spectrum is “VIBGYOR”. Violet colour deviates through maxi. Value and red colour deviates through the minimum angle. Causes of Dispersion: Each colour has in own wavelength according to Cauchy’s formula R.I. of a material depends on wavelength ( λ) B C µ = A + 2 + 4 + .......... λ λ

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For a prism of small refracting angle of deviation is Angular dispersion: It is the angle in which all colous of white light are contained dv = deviation of violet colour dr = deviation of red colour (θ) ∴ Angular dispersion = dv - dr dV = ( µV − 1) A As d S = ( µV −1) A ∴ dV − d r = ( µV −1) A − ( µr −1) A = ( µV −1 − µr +1) A dV − d r = ( µV − µ r ) A Dispersive Power (w): Ratio of angular deviation to the mean position produced by the prism. d −d ( µ − µ r ) = dµ ω= v r = v ( µ − 1) ( µ − 1) d Note: Single prism produces both deviation and dispersion simultaneously. It cannot give deviation without dispersion (or) dispersion without deviation. However a suitable combination of two prisms can do so. Dispersion of light occurs because velocity of light in a material depends upon its colour There is no dispersion of light refracted through a rectangular glass slab. Combinations of prim: I) Deviation without dispersion: Net dispersion = 0, Net deviation ≠ 0 Necessary condition ( δV − δ R ) + δ V1 − δ R1 = 0

( µV

(

(

)

− µ R ) A + µV' − µ

' R

) A'= 0

In this Situation

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 µ − µR + Net deviation = δ1 + δ ' = ( µ − 1) + ( µ '−1) A ' = ( µ − 1) A + µ − 1  − V' '  µV − µ R

(

= ( µ − 1) A − ( µ '− 1)

( µV − µ R )

(µ

− µ R'

' V

)

)

  

A

 ( µ − µ R ) ( µ '− 1)  = ( µ − 1) A V × '  µ V − µ R'   ( µ − 1)

(

)

ω   ω1  Usually ω'