Form 4 Physics Chapter 5

Physics Module Form 4

5.1

Chapter 5 - Light

GCKL 2011

UNDERSTANDING REFLECTION OF LIGHT

What light is?

Is a form energy. Light travel in a straight line and high speed about 300,000 km s-1.

How the light ray reflected by the surface of mirror?

1. The light ray that strikes the surface of the mirror is called incident ray. 2. The light ray that bounces off from the surface of the mirror is called reflected ray. 3. The normal is a line perpendicular to the mirror surface where the reflection occurs. 4. The angle between the incident ray and the normal is called the angle of incidence ,i. 5. The angle between the reflected ray and the normal is called the angle of reflection, r.

What is the Law of Reflection ? Draw the ray diagram of the plane mirror

AO = incident ray OB = reflected ray i = angle of incident r = angle of reflected

The Laws Of Reflection 1. The incident ray, the reflected ray and the normal all lie in the same plane The angle of incident, i, is ….equal….. to the angle of reflection, r.

1. Consider an object O placed in front of a plane mirror. 2. Measure the distance between the object o and the mirror. 3. Measure the same distance behind the mirror and mark the position as the image. 4. Draw the diverging ray from a point on the image to the corner of the eye. The rays from the image to the mirror must be dotted to show that are virtual. 5. Finally, draw two diverging rays from the object to the mirror to meet the diverging rays from the image.

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Physics Module Form 4 State the characteristics of the image formed by plane mirror

What is meant by virtual image? What is meant by real image?

Chapter 5 - Light

GCKL 2011

1 laterally inverted 2. same size as the object 3. virtual 4. upright 5.distance between image and mirror same as distance between object and mirror.

Image that …cannot………. be seen on a screen.

Image that …...can…be seen on a screen. CURVED MIRRORS: Concave mirror

Convex mirror

f

f

r

r 1.Light (diverged, converged) 2. (virtual,real) principal focus 3. PF= ….Focal length… = Distance between the real principal focus and the pole of the mirror.

State the differences between concave mirror and convex mirror

1.Light (diverged, converged) 2.(virtual,real) principal focus 3.PF = Focal length = Distance between the virtual principal focus and the pole of the mirror.

Common terminology of reflection of light on a curved mirror

Refer to the diagrams above and give the names for the following: 1.Centre of curvature ,C = The geometric centre of a hollow sphere of which the concave or convex mirror is a part. 2.Pole of mirror, P = The centre point on the curved mirror 3.Radius of curvature ,r = CP = radius of the curvature 4.Focal length, f = The distance between the principle focus, F and the pole of the mirror, P 5.Object distance, u = Distance of object from the pole of the mirror, P 6.Object distance , v = Distance of image from the pole of the mirror,P

Construction Rules for Concave Mirror Ray 1

Ray 2

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Ray 3

Physics Module Form 4

Chapter 5 - Light

A ray parallel to the principle axis is reflected to pass through F. Image formed by concave mirror:

A ray through F is reflected parallel to the principle axis.

GCKL 2011 A ray through C is reflected back along its own path.

Using the principles of construction of ray diagram, complete the ray diagrams for each of the cases shown below: u = object distance; v = image distance ; f = focal length ; r = radius of curvature Note: Point of intersection in the position of the image A

u < f ( Object between F and P )

Characteristics of image: 1.virtual 2.upright 3.magnified

Application: 1.magnifying mirror 2.sharing mirror 3. make-up mirror

B

u = f ( Object, O is at F )

Characteristics of image: 1.Image at infinity

Application: A reflector to produce parallel beam of light such as a reflector in 1. torchlight 2.spotlight

C

f < u < 2f or f < u < r ( Object O is between F and C

Characteristics of image: 1.magnified 2.real 3.inverted

I

D

u = 2f or u = r ( Object ,O is at C)

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Characteristics of image: 1.same size 2.real 3.inverted

Application: 1.reflector in a projector

Physics Module Form 4

Chapter 5 - Light

Eu > 2f or u > r ( Object, O is beyond C )

GCKL 2011

Characteristics of image: 1.diminished 2.real 3.inverted

I

F u =  ( Object ,O very far from the lens)

Characteristics of image: 1.diminished 2.real 3.inverted

Application: Used to view distant objects as in a reflecting telescope

I

Construction Rules for Concave Mirror Ray 1 A ray parallel to the principal axis is reflected as if it came from F. Image formed by concave mirror:

Ray 2 A ray towards F is reflected parallel to the principal axis.

Ray 3 A ray towards C is reflected back along its own path.

Using the principles of construction of ray diagram, complete the ray diagrams for each of the cases shown below: u = object distance; v = image distance ; f = focal length ; r = radius of curvature A u < f ( Object between F and P )

Characteristics of image: 1.diminished 2.virtual 3.upright

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Application: 1. Blind Conner mirror 2.Wide side view mirror

Physics Module Form 4

Chapter 5 - Light

GCKL 2011

Check Yourself: Objective Question: 1. Which of the following is true of the laws of reflection f light? A The angle of incident is equal to the angle of refraction B The incident ray and the reflected ray are always perpendicular to each other. C The incident ray , the reflected ray and the normal line through the point of incidence, all lie on the same plane.

4.

A boy stands in front of a plane mirror a distance 5 m . When the boy moves toward the mirror by 2 m , what is the distance between the boy and his new image? A C E

5. 2. The diagram shows a single ray of light being directed at a plane mirror.

What are the angles of incidence and reflection? Angle of incidence Angle of reflection A 40o 40o B 40o 50o C 50o 40o o D 50 50o

virtual smaller bigger three times as far away

A light ray incident onto a plane mirror at an angle of 50o The characteristics of an image , formed by a convex mirror for all positions of the object are A diminished, real and inverted B magnified , real, and upright C diminished ,virtual and upright D magnified , virtual and inverted

7.

A concave mirror has a focal length 20 cm. What happen to the size of image when an object is placed at a distance of 40 cm in front of the mirror? A B C

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4m 8m

6.

3. The diagram shows a ray of light from a small bulb strikes a plane mirror.

Where is the image of the bulb formed and its characteristic? A At P and virtual B At Q and real C At R and virtual

B D

An object is placed in front of a plane mirror. Compare to the object, the image formed in the mirror is always A B C D

40°

2m 6m 10 m

diminished magnified same size of object

Physics Module Form 4 8.

Chapter 5 - Light

The figure shows a candle placed in front of a concave mirror of focal length, f.

GCKL 2011

Section A (Paper 2) Structure Question: 1. Diagram 3.1 shows a mirror at the corner of a shop.

The image formed is A real, upright and magnified B real, inverted and diminished C virtual, inverted and magnified D virtual, upright and diminished 9.

DIAGRAM 3.1 / RAJAH 3.1 (a) Name the type of mirror shown in Diagram 3.1 Convex mirror …………………………………………………….. [1 mark]

When an object is placed at a point 20 cm in front of a concave mirror, a real image of the same as the object is formed on a screen placed next to the object. What is the focal length of the mirror? A B C D

(b) Name one characteristic of the image formed by the mirror. Upright / diminished / smaller / virtual …………………………………………………….. [1 mark] (c) Sketch a ray diagram to show how the image is formed. 1. Draw a parallel ray from the object that is incident along a path parallel to the principal axis appears to go through the focal point

5 cm 10 cm 15 cm 20 cm

10. Which of the following states the right reason for replacing a plane mirror are used as rear- view mirrors in motor vehicles with a convex mirror ? A B C D

2. A radial ray that is incident through the centre of curvature, C of the curved mirror is reflected back along the incident path through point C

To shine the object To widen the field of view To produce a brighter image To produce a sharper image

3. Determine the correct position of the image

Answer: 1 2 3 4 5 6 7 8 9 10

C D A C A C A A B C

[3 marks] (d) What is the advantage of using this type of mirror in the shop? To increase the field of vision …………………………………………………………… [1 mark] 5-6

Physics Module Form 4

Chapter 5 - Light

2. Diagram 4.1 shows the image of a patient’s teeth seen in a mirror used by a dentist.

GCKL 2011

Section B(Paper 2) Essay Question(20 marks) Diagram 7.1 shows two cars, P and Q , travelling in the opposite directions, passing through a sharp band. A mirror is placed at X .

DIAGRAM 4.1 (a) Name the type of the mirror used by the dentist. Concave mirror ………………………………………… ……. [ 1 mark ] DIAGAM 7.1

(b) State the light phenomenon that causes the image of the teeth

(a) Diagram 7.2 shows an incomplete ray diagram when a plane mirror is placed at X.

Reflection of light ……………………………………………….......... [ 1 mark ] (c) State two characteristics of the image formed. Virtual, upright and magnified ……………………………………………………. [ 2 marks ] (d) In the diagram below, the arrow represents the teeth as the object of the mirror. Complete the ray diagram by drawing the required rays to locate the position of the image.

DIAGRAM 7.2 (i)

Complete the ray diagram in Diagram 7.2 [2 marks]

1. Two reflected rays are shown (diagram)[1 mark] 2. Angle of incidence = Angle of reflection (diagram) [1 mark]

[ 3 marks] 5-7

Physics Module Form 4

Chapter 5 - Light

GCKL 2011

(ii) State the light phenomenon involved in (a)(i). (ii) Reflection ………………………………………………… [1 mark]

Complete the ray diagram in Diagram 7.3 [2 marks]

(iii) Based on your answer in (a)(i), state the problem experienced by the driver in car P. The driver in car P cannot see car Q // field of ……………………………………………………….. view very small ……………………………………………………….. [1 mark] 1. Two reflected rays are shown (diagram)[1 mark] 2. Angle of incidence = Angle of reflection (diagram) [1 mark]

(b) Diagram 7.3 shows an incomplete ray diagram when a curve mirror is placed at X to replace the plane mirror in Diagram 7.2. The curve mirror is used to overcome the problem that occur in (a)(iii).

(iii) Based on your answer in b(ii), how the curve mirror solved the problem in (a) (iii)? The convex mirror increase the field of view ……………………………………………………. [1 mark]

( C) The characteristics of the image formed by the curved mirror in Figure 7.3 is diminished, virtual and upright. (i) What happen to the characteristics of the image when the focal length of the curved mirror is increased? The driver in car P cannot see car Q // field of ……………………………………………………….. view very small ……………………………………………….. [1 mark] (ii) Give the reason for your answer in (c)(i).

DIAGRAM 7.3 (i)

Give the name of the curve mirror. Convex mirror …………………………………[1 mark]

The characteristics of image of a convex …………………………………………………….. mirror not depends on the focal length ………………………………………………… [1 mark]

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Physics Module Form 4

5.2

Chapter 5 - Light

GCKL 2010

UNDERSTANDING REFRACTION OF LIGHT

The diagram shows the spoon bent when put inside the water. State the phenomenon occurs.

Refraction of light

How the phenomenon occurs?

Light travel from less dense medium which is air to denser medium (water), light will be deviated near to the normal. Thus the spoon seems like bending after putting inside the water.

Why light is refracted?

It due to change in the velocity of light as it passes from one medium into another. Light travel more slowly in water (or glass) than in air. When a light beam passes from air into glass, one side of the beam is slowed before the other. This makes the beam ‘bend’.

Three different cases of refraction

Case 1:

Case 2:

Case 3:

i = 0 ,r = 0

i>r

i Refractive index of liquid L ]

What is the refractive index of medium X? A 0.85 B 1.24 C 1.31 D 1.41 E 1.58 11 The diagram shows a light ray travels from the oil into the air.

8

The diagram shows a light ray which travels from the air to the glass.

What is the value of k? [ Refractive index of oil = 1.4 ] A B C D E

What is the refrective index of the glass? A

C

Sin S Sin Q

B

Sin Q Sin R

D

Sin P Sin R

44.4o 45.6o 54.5o 55.4o 58.9o

12 The diagram shows a light of ray travels from the air into a glass block.

Sin R Sin S

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Physics Module Form 4

Chapter 5 - Light

GCKL 2010

15 The diagram shows a coin is put at the base of the beaker. The image of the coin appears to be 5 cm from the base of the beaker.

What is the refractive index of the glass block? A B C D E

1.38 1.45 1.51 1.62 1.74

A C

13 The speed of light in the air is 3 x 108 ms-1 . What is The speed of light in a plastic block? [ Refractive index of plastic = 1.2 ] A B C D E

E

1.0 x 108 ms-1 1.5 x 108 ms-1 2.0 x 108 ms-1 2.5 x 108 ms-1 3.0 x 108 ms-1

Answer: 1 2 3 4 5 6 7 8 9 10

14 The diagram shows a boy appearing shorter when he is in a swimming pool. The depth of the water in the pool is 1.2 m. [ Refractive index of water = 1.33 ]

What is the apparent depth of the pool? A C E

0.1 m 0.9 m 1.6 m

What is the refractive index of the liquid? 8 B 5 13 8 11 D 13 5 8 19 14

B D

0.3 m 1.1 m 5 - 16

D B D A D C B C D A

11 12 13 14 15 16 17 18 19 20

A C D C D

Physics Module Form 4

Chapter 5 - Light

GCKL 2010

Section A (Paper 2) Structure Question: (C ) (i) Draw a Diagram of the light ray shown on diagram 3.1, meeting the water surface RS, and show its path after meeting the surface. [1 mark]

1. The Diagram shows a side view of a water-filled aquarium RSTU. An electric lamp, surrounded by a shield with a narrow transparent slit, is immersed in one corner of the aquarium at U. The light ray from the slit shines on the water surface RS at an angle of 40o as shown in diagram below.

R

Water

40o

40o

S

S

R 40

Water

o

Aquarium

Light ray

Light ray U

T

T

U DIAGRAM 3.1

(a)

ii. Calculate the angle that this new path makes with RS and label the angle. [2 ma [1 mark] Angle = 40o

What is meant by refractive index of a substance?

(d) The lamp is then placed outside underneath the aquarium with the light striking to the bottom of the aquarium as shown in Diagram 3.2. Draw the light ray on Diagram 3.2, after striking the aquarium.

Refractive index is an indication of the light bending ability of the medium / n = sin i sin r

[1 mark]

[1 mark]

(b) If the refractive index of water is 1.33, calculate the critical angle for a ray travelling from water to air. water n = 1 sin c sin c = 1 1.33 c = 48.80

[ 2 marks] Light ray Lamp

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Physics Module Form 4

Chapter 5 - Light

2. An observer is looking at a piece of coin at the bottom of a glass filled with water as shown in Diagram 3. He found that the image of the coin is nearer to the surface of the water.

3.

GCKL 2010

Figure(a) shows an object in a small pond. The depth of the water in the pond is H. The image of the objet appears to be h from water surface.

[ 2 m a r k s ]

Figure(a)

(a) State the relationship between H and h (a)(i)

State a characteristic of image in Diagram 3. Virtual/magnified

(ii)

When H increases, h increases/ H is directly proportional to h ....................................................................

[1 mark]

(b) When H = 4.5 ]m and the refractive index of water is 1.33, determine the value of h .

Name the science phenomenon involve in the observation above. [1 mark] Refraction

n 1.33

(b)

Explain why the image of the coin appears nearer to the surface of the water. - Light ray travels from density to less density medium - Refracted ray away from normal

(c) What happen to value of h when the pond is poured with water of refractive index 1.40 ?

[2 marks] (c)

H decreases ……………………………………………

On Diagram 3, complete the ray diagram from the coin to the observer's eye. [2 marks] -Draw

= Real depth , H Apparent depth, h = 4.5 m H H = 3.38 m

refracted ray correctly

- Draw ray from image to the observer [

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Physics Module Form 4

Chapter 5 - Light

GCKL 2010

Section B (Paper 2) Essay Question ii. Observe Figure 4(a) and Figure 4(b) carefully. Compare the common characteristics of the pencil and the print before and after they are removed from the water and the glass block respectively. Use a physics concept to explain the appearance of the pencil and the print in water and under the glass block respectively. [5 marks]

1. Figure 4(a) shows a pencil placed in a glass of water. Figure 4(b) shows the appearance of print viewed from the top of a thick block of glass placed over it. pencil

Glass block

Answer: 1. The pencil appears bent when placed in water and the print appears raised when a thick block of glass is placed over it. water Figure 4(a)

2. The rays of light from the pencil are refracted away from the normal as they leave the water and enter the eye of the observer. These rays appear to come from a virtual image above the actual point. The pencil ,therefore , appears bent in the water.

Figure 4(b)

(a) i. Why does the pencil appear bent to our eyes? Why does the print appear raised? [1 mark]

3. Rays of light from the print below the glass are refracted away from the normal as they leave the glass and enter the aye of the observer . The writing, therefore, appears to be slightly raised.

Answer: We can see the pencil and the print because the rays of light from the two objects reach our eyes.

4. Refraction of light is the physics concept involved. 5. Refraction of light is a phenomenon in which rays of light change direction when they pass from one medium to another medium of a different density.

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Physics Module Form 4

5.3

What is meant by total internal reflection?

Chapter 5 - Light

GCKL 2011

UNDERSTANDING TOTAL INTERNAL REFLECTION

Total internal reflection is the complete reflection of light ray travelling from a denser medium to a less dense medium. Total: because 100% of light is reflected Internal: because it happens inside the glass or denser medium.

What is meant by critical angle ,c?

The critical angle, c, is defined as the angle of incidence (in the denser medium) when the angle of refraction (in the denser medium), r is 90°.

What are the relationship between the critical angle and total internal reflection ?

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Physics Module Form 4 

Chapter 5 - Light

When the angle of incidence, i keeps on increasing, r too increases



And the refracted ray moves further away from the normal



And thus approaches the glass- air boundary.



 



State the two conditions for total internal reflection to occur

GCKL 2011

The refracted ray travels along the glass-air boundary. Angle of refraction, r = 90°. This is the limit of the light ray that can be refracted in air as the refracted in air cannot be any larger than 90°. The angle of incidence in the denser medium at the limit is called the critical angle, c.





If the angle of incidence is increased is increased further so that it is greater than the critical angle, (i > c): - no refraction - all the light is totally in the glass This phenomenon is called total internal reflection.

1. light ray enters from a denser medium towards a less dense medium. 2. the angle of incidence in the denser medium is greater then the critical angle of the medium ( i > c)

What are the relationship between the refractive index, n and critical angle, c?

What are the phenomena involving total internal reflection?

1. Mirage

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

In hot days, a person traveling in a car will see an imaginary pool of water appearing on the surface of the road.



The layes higher up are cooler and denser.



Light ray from the sky travels from denser to less dense medium and

Physics Module Form 4

Chapter 5 - Light

GCKL 2011 will refracted away from the normal. 

The angle of incidence increases until it reach an angle greater than the critical angle.



Total internal reflection occurs and the light is reflected towards the aye of the observer.



If the observer’s eye is in the correct position, he will see a pool of water(image of the sky) appearing on the road surface.



This is known as a mirage.



When sunlight shines on millions of water droplets in the air after rain, a multi coloured arc can be seen.



When white light from the sun enters the raindrops, it is refracted and dispersed into its various colour components inside the raindrops.



When the dispersed light hit the back of the raindrop, it undergoes total internal reflection.



It is then refracted again as it leaves the drop.



The colours of a rainbow run from violet along the lower part of the spectrum to red along the upper part.

2. Rainbow

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Physics Module Form 4 Give some examples of application of total internal reflection.

Chapter 5 - Light

1. The sparkling of a diamond

2. Periscope

GCKL 2011 

A diamond has a high refractive index.



The higher the refractive index, the smaller the critical angle.



A small critical angle means total internal reflection readily occurs.



Light is easily reflected inside the diamond.



In this way, more light will be confined within the diamond before refracting out into the air.



The periscope is built using two right angled 45° made of glass. The critical angle of the prism is 42°.



The angle of incidence is 45° which is greater than the critical angle.



Total internal reflection occurs.



The characteristics of the image are: Virtual, upright, same size.

Give the advantages of the prism periscope compared to mirror periscope. Answer: (i) The image is brighter because all the light energy reflected. (ii) The image is clearer because there are no multiple images as formed in a mirror periscope.

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Physics Module Form 4

Chapter 5 - Light

3. Prism Binocular

GCKL 2011 

A light ray experiences two total internal reflections at each prism.



So the final image in binoculars is virtual, upright and same size.

What are the benefits of using prism in binoculars? (a) an upright image is produced. (b) The distance between the objective lens and the eyepiece is reduced. This make the binoculars shorter as compared to a telescope which has the same magnifying power.

4. Optical Fibres



The external wall of a fibre optic is less dense than the internal wall.



When light rays travel from a denser internal wall to a less dense external walls at an angle greater than the critical angle, total internal reflection occurs.

Give the advantage of using optical fibres cables over copper cables. (1) they are much thinner and lighter. (2) a large number of signals with very little loss over great distances. (3) The signals are safe and free of electrical interference

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Physics Module Form 4

Chapter 5 - Light

GCKL 2011

(4) It can carry data for computer and TV programmes.

Check Yourself: Objective Question: 1

A ray of red light travelling in glass strikes the glass-air boundary . Some light is reflected and some is refracted. Which diagram shows the paths of the rays?

[ Refractive index of medium X = 1.3 Refractive index of medium Y = 1.5 ]

4

Which of the following shows total internal reflection?

5

The diagram shows light ray XO experiencing total internal reflection when travelling from the glass to air.

2 One of the diagram below shows the path of a beam of light that is incident on a water-air surface with angle of incidence greater than the critical angle. Which one is it?

3

Which of the following diagram correctly shows the total internal reflection of ray of light?

Which statements about total internal reflection are correct? 5 - 25

Physics Module Form 4

Chapter 5 - Light

P -  is more than the critical angle of glass Q - The speed of light in the glass is higher than in air R - The refractive index of glass is greater than air A P and Q B P and R C Q and R D P,Q and R 6

In which direction does the light move from ? A B C D

The diagram shows a semi-circular plastic block is placed in a liquid. 9

Which of the following is correct? A B C D

7

OQ OR OS OT

A ray of light incident on one side of a rectangular glass block. If the angle of refraction in the glass block is 40o , which one of the following diagrams best represents this ray? [ The critical angle of glass is 42o ]

Density of the plastic block is less than density of the liquid Refractive index of the plastic block is less than refractive index of the liquid Critical angle of the plastic block is less than critical of the liquid Angle of incidence is less than critical angle of the liquid

The diagram shows a ray of light passing through medium M to medium N.

10 The diagram shows a light ray, P, directed into a glass block. The critical angle of the glass is 42o. In which direction does the light move from point Q?

Which of the following is correct? A The angle of reflection is 55o B The critical angle of medium M less than 35o C Density of medium M less than the density of medium N 8

GCKL 2011

The figure shows a ray of light PO traveling in a liquid strikes the liquid-air boundary. [ The critical angle of the liquid = 45o ] 5 - 26

Physics Module Form 4

Chapter 5 - Light

GCKL 2011 C D

the greatest angle of incidence in optically more dense medium the greatest angle of incidence in optically less dense medium

14 Which of the following shows the correct critical angle , c of the semi- circular glass block ?

11 The diagram shows a light ray , M, directed into a glass block. The critical angle of the glass is 42o. In which diagram does the light move from point O?

15 The diagram shows a light ray travelling from air into a plastic block with an angle of incidence ,X. What is the critical angle of the plastic?

12 The figure shows a ray of light is incident in air to the surface of Prism A and B.

16 The diagram shows a light ray travelling from air into a glass prism. Which comparison is correct ? A

Density of prism A < density of prism

B B C

13

Critical angle of prism A < critical angle of prism B Refractive index of prism A < refractive index of prism B The critical angle is What is the critical angle of the glass?

A B

the smallest angle of incidence in optically more dense medium the smallest angle of incidence in optically less dense medium 5 - 27

Physics Module Form 4

Chapter 5 - Light

A 40o C 60o E 80o

B 50o D 70o

GCKL 2011

21 The diagram shows a cross- section of a fibre optic cable.

17 The refractive index of water is 1.33. What is the critical angle of the water. A 44.5o B 46.9o o C 48.8 D 49.2o E 54.3o 18 The refractive index of plastic block is 13 . 5 What is the value of the cosine of the critical angle of the plastic? A

5 12

B

12 13

C

13 12

D

5 13

E

13 5

Which comparison is correct ? A B C

Answer: 1 2 3 4 5 6 7 8 9 10

19 The figure shows a ray of light AO traveling in medium X strikes the medium X-air boundary. [ The refractive index of medium X = 1.12 ]

In which direction does the light move from O ? A C

OE OC

B D

OD OB

20 Which of the following not applies the principle of total internal reflection? A B C D

Prism binocular Mirror periscope Optical fibre Road mirage 5 - 28

Density of P < density of Q Density of P >density of Q Density of P = density of Q

A D D D B C B D A C

11 12 13 14 15 16 17 18 19 20

C B C C D B C D C A

Physics Module Form 4

Chapter 5 - Light

GCKL 2011 (c)

Section A (Paper 2) Structure Question:

Name other optical device that applies the phenomenon in (a)(i). [ 1 mark ]

Prism periscope // prism binoculars // camera// endoscope and etc.

1. Diagram 1 shows a cross-sectional area of an optical fibre which consist of two layers of glass with different refractive index. The glass which forms the inner core, Y is surrounded by another type of glass which forms the outer layer, X.

2. Figure 4 shows a traveller driving a car on a hot day. The traveller sees a puddle of water on the road a short distance ahead of him. Puddle of water

Figure 4 DIAGRAM 1 (a) (i) Name the light phenomenon observed in optical fibre?

a) Which part of the air is denser? Close to the sky / cool air ………………………………………………………… ( 1 mark )

Total internal reflection

[ 1 mark ] (ii) Compare the refractive index of outer layer X and inner core Y. The refractive index of Y is higher than the refractive index of X// Vice versa

b) Name a phenomena of light that always depends on the air density when light travels from the sky to the earth before it reaches point X. Refraction ………………………………………………………… ( 1 mark )

[ 1 mark ]

c) i) What is the phenomenon occurring at point X (b) The refractive index of inner core Y is 2.10. Calculate the critical angle of the inner core Y. Sin c =

Total internal reflection ……………………………………………………..… (1 mark )

1 1 = = 0.4762 n 2 .1

ii)

c = 28.44o // 280 26’

What is the puddle of water actually?

The image of sky ………………………………………………………… ( 1 mark )

[2 marks] 5 - 29

Physics Module Form 4

Chapter 5 - Light

GCKL 2011

d) Using the diagram above, explain how the traveller can see the puddle of water on the road. 1.Light from sky to the earth refracted 2. The light reach at a point X, total internal reflection occurred ………………………………………………………… ( 2 marks ) e) Name one optical instrument that uses the phenomenon in (d) Optical fibre ………………………………………………………… ( 1 mark )

Answer: Glass prism

Object 45o

3. Completing the ray diagram below, to show how a periscope works: (critical angle of glass = 42o)

object

tctct

Glass prism

Total internal reflection takes place because angle of incident > critical angle

Eye Eye e

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Physics Module Form 4

5.4

Chapter 5 - Light

GCKL 2011

U N D E R S T A N D I N G

L E N S E S

Introduction Lenses are made of transparent material such as glass or clear plastics. They have two faces, of which at least one is curved. Types of lenses

State the differences between convex lens and concave lens

(a) Convex lens, also known as converging lens. It is thicker at the centre of the lens.

Convex lens

Concave lens

When light ray which are parallel and close to the principle axis strikes on a convex lens, they are refracted and converge to a point, F on the principle axis. This point is a focal point of the convex lens. Common terminology of reflection of light on a curved mirror

(b) Concave lens, also known as diverging lens. It is thinner at the centre of the lens.

When light rays are parallel to the principle axis fall on a concave lens., they are refracted and appear to diverge from the focal point on the principle axis.

1. The focal point, F is a point on the principle axis where all rays are close and parallel to the axis that converge to it after passing through a convex lens, or appear to diverge from it after passing through a concave lens. 2. The focal length, f is the distance between the focal point and the optical centre. 3. The optical centre, C is the geometric centre of the lens. It is the point through which light rays pass through without deviation. 4. The principle axis is the line passing through the optical centre, C. 5 - 31

Physics Module Form 4

Chapter 5 - Light

GCKL 2011

Construction rules of convex lens

Rule 1: A ray parallel to the principle axis is refracted through the focal point, f. Rule 2: A ray passing through the focal point is refracted parallel to the principle axis. Rule 3: A ray passing through the optical centre, C travels straight without bending. The point of intersection is the position of the image. The images formed by a convex lens depend on the object distance, u.

Images form by convex lens

Using the principles of construction of ray diagram, complete the ray diagrams for each of the cases shown below: u = object distance; v = image distance ; f = focal length Note: Point of intersection in the position of the image A

u < f ( Object between F and P )

Characteristics of image: 1.virtual 2.upright 3.magnified 4.Same side as the object

Application: 1.magnifying glass spectacle 2.lens for longsightedness.

B

u = f ( Object, O is at F )

Characteristics of image: 1.virtual 2.upright 3.magnified 4. Same side as the object

Application: 1. to produce a parallel a parallel beam of light , as in a spotlight, astronomical telescope

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Physics Module Form 4

Chapter 5 - Light

GCKL 2011

C f < u < 2f or f < u < r ( Object O is between F and C

Characteristics of image: 1.real 2.inverted 3.magnified 4. On apposite side of the object.

Application: 1.projector lens 2.photograph 3. enlarger 4.objective lens of microscope

D

Characteristics of image: 1.real 2.inverted 3.same size as the object 4. On the opposite side of the object

Application: 1.photocopying machine

Eu > 2f or u > r ( Object, O is beyond C )

Characteristics of image: 1.real 2.inverted 3.magnified

Application: 1.magnifying mirror 2.sharing mirror 3. make-up mirror

F u =  ( Object ,O very far from the lens)

Characteristics of image: 1.virtual 2.upright 3.magnified

Application: 1.magnifying mirror 2.sharing mirror 3. make-up mirror

u = 2f or u = r ( Object ,O is at C)

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Physics Module Form 4

Chapter 5 - Light

Construction rules of concave lens

GCKL 2011

Ray 1: A ray parallel to the principle axis is refracted as if it appears coming from the focal point, F which is located at the same side of the incident ray.

1

2

Ray 2: A ray passing through the focal point is refracted parallel to the principle axis.

3

Ray 3: A ray passing through the optical centre, C travels straight without bending. The point of intersection is the position of the image . The image formed by a concave lens are always : Virtual, upright and diminished. Image formed by convex mirror:

Using the principles of construction of ray diagram, complete the ray diagrams for each of the cases shown below: u = object distance; v = image distance ; f = focal length A u < f ( Object between F and P )

Characteristics of image: 1.diminished 2.virtual 3.upright

Application: 1. Blind Conner mirror 2.Wide side view mirror

(B ) f fe

fe > fo

Final image , virtual image at infinity

Final image , virtual image, inverted , magnified

The diagram shows the arrangement of the lenses in a telescope. The power of lens A and lens B are 0.5 D and 5D respectively.

(a) Why do the light rays PQ and RS are parallel? (1 mark)

(a) Complete the ray diagram in the diagram above. Answer:

Because the object is from infinity …………………………………………. (b) State the focal length of (i) lens A (2 marks) P = 1 , fo= 1 = 2.0 cm fo 0.5 (ii)

lens B (2 marks) P = 1 , fe= 1 = 0.2 cm fe 5 (c) What is the distance between lens A and lens B? (2marks)

(b) State the characteristics of the final image formed. (1 mark) Inverted, virtual and magnified ..................................................................

fo + fe = 2.0 + 0.2 = 2.2 m 5 - 47

Physics Module Form 4

Chapter 5 - Light

(d) What is the linear magnification of the telescope? (2 marks) m = fo, = 2.0 = 10 fe 0.2 (e) In the diagram above , complete the ray diagram of the telescope. (3 marks) (f) State the characteristics of the final image formed. (2 marks) Virtual , formed at infinity, inverted ...................................................................

3

The figure shows the lens and mirror arrangement for a slide projector.

(a)

What is the function of (i) concave mirror (1 mark) Reflects light to the condenser .......................................................... (ii)

condenser lens (1 mark) Acts as a heat filter to protect slide from getting overheated. .......................................................... (b) State two normal adjustment should be done while using the slide projector.(2 marks) 1. The slide being the object is placed between f and 2f from the projector lens . .................................................................. 2. The slide should be placed up side-down in order to form an image on the upright screen. ……………………………………………… (c) State the characteristics of the final image formed.. (2 marks) Real, magnified and upright .................................................................. 5 - 48

GCKL 2011