Chapter18pp161-170

OCR (A) specifications: 5.3.1a,b,c,d,e,f

Chapter 18 Reflection and refraction Worksheet Worked examples Practical 1: Demonstrating refraction and total internal reflection Practical 2: Snell’s law and refractive index End-of-chapter test Marking scheme: Worksheet Marking scheme: End-of-chapter test

Worksheet Intermediate level 1 2 3 4

State the two laws of reflection.

[2]

Explain what is meant by refraction of light.

[1]

Define refractive index n of a transparent medium.

[1]

The diagram shows rays of light entering two transparent materials A and B. 45°

Explain which material has a greater refractive index. [1]

5

6

45°

air

air

A

B

The speed of light in turpentine is 2.03 × 108 m s–1. a

Calculate the refractive index of turpentine.

[3]

b

A ray of light in air travels towards the air/turpentine interface. The angle of incidence is 60°. Determine the angle of refraction in the turpentine.

[3]

The refractive index of ruby is 1.76. Calculate the critical angle for ruby.

[3]

Higher level 7

The diagram shows the path of light from a laser as it passes through a tank of water. a

Determine the angle of incidence of the ray of light at the interface XY. [1]

b

Calculate the angle θ. (The refractive index of water is 1.33.) [3]

θ

water N

X

Z

Y

46°

laser light

8

A section through an optic fibre is shown below. 200 m

input signal

glass

output signal

The fibre is made of a material of refractive index 1.54. a

Calculate the critical angle for the fibre.

[2]

b

Calculate the time taken for a ray of light to travel the length of the fibre when travelling along its axis.

[2]

Explain why the output signal from the fibre is ‘smeared’ and not as sharp as the input signal.

[1]

c 162

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18 Reflection and refraction

Extension 9

The diagram shows white light incident at one face of a glass prism.

60°

The refractive index of the glass is different for different wavelengths of visible light. For blue light the refractive index is 1.521 and for red light the refractive index is 1.510. Calculate the ‘divergence’ angle δ between the blue and red light emerging from the second face of the prism. [7]

90° 70° 90°

white light

δ

prism 60°

red

60° blue

Total: ––– Score: % 30

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163

Worked examples Example 1 A ray of light enters a block of ice at an angle of incidence of 60°. Calculate the angle of refraction for light in the ice. The refractive index of ice is 1.31. Snell’s law: n = i = 60° 1.31 =

sin i sin r

Remember that this equation is valid for light travelling from vacuum (or air) into the material.

r=? sin 60° , sin r

sin r =

sin 60° = 0.6611 1.31

r = sin–1 (0.6611) ≈ 41.4°

Example 2 The diagram shows light travelling from water into ice. 90°

ice water

Calculate the critical angle for the water/ice interface. (Refractive index of ice = 1.31; refractive index of water = 1.33.)

C

You can use Snell’s law: n=

weak reflection

sin i sin r

but you need to be aware that n is the (absolute) refractive index of the material when light travels from a vacuum into the material. In this example we need to find the refractive index n for the water/ice interface for light travelling from ice into water. Hence: n=

speed of light in ice speed of light in water 3.0 × 108

n=

3.0 × 10

sin C =

1.31

8

1.33

=

1.33 = 1.015 1.31

1 1 = = 0.9850 n 1.015

therefore C = 80.1°

Tip You can use a useful ‘variation’ of Snell’s law. For a ray of light incident on or refracted at an interface: n × sin (angle) = constant For the incident light in water, we have: n = 1.33 and

angle = C (the critical angle)

For light refracted into ice, we have:

angle = 90°

n = 1.31 and

Hence: 1.33 × sin C = 1.31 × sin 90° C = sin–1 (1.31/1.33) = 80.1°

164

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18 Reflection and refraction

Practical 1 Demonstrating refraction and total internal reflection Safety When using a laser pen, do not look directly into the beam. Laser light will permanently damage your eyes. It is essential to wear laser safety goggles. Also be careful of reflected laser light. The room does not need to be very dark for this experiment. Teachers and technicians should follow their school and departmental safety policies and should ensure that the employer’s risk assessment has been carried out before undertaking any practical work.

Apparatus • • • •

laser pen (or a ray box and power supply) large transparent rectangular container few drops of milk laser safety goggles

Introduction In this experiment you have the chance to investigate what happens to the path of light as it travels from air into water and when it travels in the opposite direction.

Procedure This experiment is a variation of the one illustrated in figure 18.10 and described on page 157 of Physics 1.

1 2 3

Fill the container with water. Add just a few drops of milk into the water and stir well. When you shine the light into the water, you can see its path in the water. (The milk in the water scatters the light in all directions.)

4

Shine the light at different angles and observe the refraction (bending) of light at the water/air interface.

5

You should be able to observe total internal reflection with this arrangement.

Guidance for teachers This experiment is best done as a teacher demonstration. Doing the experiment in a semi-darkened room ensures that the pupils of the eyes are not fully dilated. You need about four drops of milk in a litre of water. For your reference, the critical angle for the water/air interface is about 49°.

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165

Practical 2 Snell’s law and refractive index Safety There are not likely to be any major hazards in carrying out this experiment. However, teachers and technicians should always refer to the departmental risk assessment before carrying out any practical work.

Apparatus • ray box • power supply for the ray box • glass or Perspex block

• protractor • plain paper

Introduction In this experiment you will determine the refractive index of glass or Perspex using Snell’s law.

Procedure Snell’s law is described on page 155 of Physics 1. The diagram shows the arrangement used to determine the refractive index of the glass.

ray box

i

r glass block

1 2 3 4

Place the glass block on the plain paper and trace out the outline of the block. Trace out the path of the incident and refracted rays for different angles of incidence. For each ray, measure the angle of incidence i and the angle of refraction r. Tabulate your results.

i (degrees)

5 6 7 8

r (degrees)

sin i

sin r

Plot a graph of sin i against sin r. Draw a straight line of best fit through the points. Determine the gradient of the line. The gradient of the line is equal to the refractive index of the glass. Explain why. What is the uncertainty in your answer for the refractive index? How does your value compare with the accepted value for the refractive index of the glass?

Guidance for teachers The arrangement above can be used to find the refractive index of water. Use a large transparent rectangular plastic box filled with water. 166

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18 Reflection and refraction

End-of-chapter test Answer all questions. speed of light in free space (vacuum) c = 3.0 × 108 m s–1

1 2

With the aid of a diagram, define Snell’s law.

[2]

a

[2]

Complete the ray diagram below to locate the position of the image.

mirror

object

b

3

The image seen by the eye is a virtual (not real) image. Explain what is meant by a virtual image. [1]

The critical angle for a particular type of glass is 43°. a b

Complete the ray diagram to show how the light emerges from the prism made of this type of glass.

[2]

Calculate the refractive index of the glass.

[3]

45°

4

45°

The diagram below shows a step-index fibre optic cable. cladding

core

cladding

The core is made of glass of refractive index 1.47 and the cladding is made of glass of refractive index 1.45. a

Calculate the speed of light in: i

the core;

[2]

ii

the cladding.

[2]

b

Calculate the refractive index for light travelling from the cladding to the core. [2]

c

Use your answer to b to determine the critical angle for the core/cladding interface.

d

Explain what is meant by multipath dispersion and state one way of minimising it. Total: ––– Score: 20

18 Reflection and refraction

[2] [2] %

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167

Marking scheme Worksheet 1

Angle of incidence = angle of reflection (angle measured relative to the normal). [1] The incident and reflected rays together with the normal lie in the same plane. [1]

2

Refraction is the bending of light as it travels from one medium into another. [1]

3

n=

4

Material A has greater refractive index because the light is deviated more towards the normal. [1]

5

a

b

speed of light in vacuum [1] speed of light in medium

n=

speed in vacuum [1] speed in medium

n=

3.00 × 108 [1]; 2.03 × 108

n=

sini [1] sinr

n=

sin 60° sinr

so

n = 1.48 [1]

sin r =

sin 60° = 0.5852 [1] 1.48

r = sin–1 (0.5852) = 35.8° [1]

6

7

sin C =

1 n

sin C =

1 = 0.5682 [1]; C = 34.6° [1] 1.76

(C = critical angle) [1]

a

Angle of incidence = 90° – 46° = 44° [1]

b

n=

sin i sin r

1.33 =

sin θ [1] sin 44°

sin θ = 1.33 × sin 44° = 0.9239 [1]

θ = 67.5° [1]

8

a

sin C =

1 = 0.6494 [1] 1.54

C = 40.5° [1] b

Speed in fibre = Time taken =

c

168

3.0 × 108 = 1.948 × 108 ≈ 1.95 × 108 m s–1 [1] 1.54

200 = 1.027 × 10–6 s ≈ 1.03 µs [1] 1.948 × 108

Light that is internally reflected travels a much longer distance than light travelling along the axis of the fibre. The light from the input pulse exits the fibre at different times. Hence the output signal is ‘smeared’ or ‘spread out’. [1]

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18 Reflection and refraction

9

Refraction at first face: blue light ⇒

1.521 =

sin 70° ; r = 38.16° [1] sin r

red light ⇒

1.510 =

sin 70° ; r = 38.49° [1] sin r

Refraction at second face: blue light ⇒ 1.521 =

sin θ b sin 21.84°

red light ⇒ 1.510 =

angle of incidence in glass = 60° – 38.16° = 21.84° [1]

θ b = 34.46° [1]

angle of incidence in glass = 60° – 38.49° = 21.51° [1]

sin θ r sin 28.51°

θ r = 33.62° [1]

δ = θ b – θ r = 34.46° – 33.62° = 0.84° [1]

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Marking scheme End-of-chapter test 1

Correct diagram with labels. [1] i vacuum medium r

Refractive index =

2

a

sini , where i = angle of incidence and r = angle of refraction [1] sin r image

Two correct rays shown. [1] Image located correctly. [1]

b

The image cannot be formed on a screen, since no light rays reach this point. [1]

mirror

object

3

a

Correct total internal reflection at each face. [1] Correct path of ray. [1]

b

4

a

1 sin C = (C = critical angle) [1] n n=

1 [1]; sin 43°

i

n=

c v

v=

n ≈ 1.47 [1]

i >C total internal reflection

45° 45° 45°

45°

ray turned through 180°

3.0 × 108 [1] 1.47

v = 2.041 × 108 m s–1 ≈ 2.04 × 108 m s–1 [1] ii

n=

c v

v=

3.0 × 108 [1] 1.45

v = 2.069 × 108 m s–1 ≈ 2.07 × 108 m s–1 [1] 2.069 × 108 [1]; 2.041 × 108

b

n=

c

sin C =

n = 1.014 ≈ 1.01 [1]

1 = 0.9862 [1] 1.014

C ≈ 80.5° [1] d

Rays of light take different times to travel the length of the optic fibre. The rays travelling along the axis of the fibre take less time than the rays that are totally internally reflected. This leads to the signal being ‘smeared’. [1] Multipath dispersion can be minimised by using a monomode cable. [1]

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