Place a rectangular glass block on a sheet of paper and draw around it.

1. Snell’s Law i r • Place a rectangular glass block on a sheet of paper and draw around it. • Draw a normal at 90° to the top surface of the block. • Shine light rays, with angles of incidence (i) of 0°, 10°, 20°….. 90°, into the block at the point where the normal meets the glass surface. Record the angle of refraction (r). Snell’s Law states that the ratio of sin(i)/sin(r) is the same for each light ray. The ratio is referred to as the ‘refractive index ‘n’ of the material. Add another column to your table headed sin(i)/sin(r). Does Snell’s law apply in this case? What is the refractive index for glass? Repeat the same experiment for Perspex and find the refractive index for Perspex.

2. i1 I1 and r1 are the angles of incidence and refraction at the point where the light ray enters the glass block. Sin i1 Sin r1 The refractive index of the block, n = r1 i2 I2 and r2 are the angles of incidence and refraction at the point where the light ray leaves the glass block. Since, i2 = r1 and r2 = i1 Then r2 Sin i1 Sin r1 Sin r2 Sin i2 = n = or This will apply whenever the light ray goes from a substance to air 1 . n Sin i2 Sin r2 =

3. A light ray is directed from glass to air. The refractive index of glass is 1.5 Calculate the angle of refraction of the light ray in air if the angle of incidence is 30º r Sin i Sin r 1 . n = i glass 1 . 1.5 Sin 30 Sin r = 1 . 1.5 Sin 30 Sin r = Sin r = 1.5 x sin30 r = 48.6º

4. Refraction by a prism r2 Q. A light ray enters an equilateral triangular prism of refractive index 1.55 at an angle of incidence of 35º . a. Calculate the angle of refraction in the glass. i2 i1 r1 Sin 35 Sin r Sin 35 1.55 1.55 = Sin r1 = r1 = 21.7º b. Calculate the angle i2 Using trigonometry, i2 = 38.3º c. Calculate the angle as the light ray leaves the prism r2 Sin i2 Sin r2 Sin i2 Sin r2 1 . n 1 . n Sin r2 = 0.96 = = r2 = 73.8

5. Explaining refraction Refraction happens because light changes speed in different substances. The smaller the speed of light in a substance, the greater its refractive index. Where: c is the speed of light in vacuum or air (3x108ms-1) cs is the speed of light in the substance c cs n = Note that the refractive index n is always >1 Since light is a wave, the wave speed equation applies: Wave speed = wavelength x frequency c = λ x f As the wave speed decreases, the wavelength decreases and the frequency stays the same. λ x f λc x f n = λ λc n =

6. Refraction at a boundary between two transparent substances When a light ray crosses a boundary from a substance in which speed of light is c1 to a substance in which the speed of light is c2: r c1. c2 Sin i Sin r = i This can be rearranged as: n2 n1 1 . c1 1 . c2 Sin i Sin r = Multiplying both sides x c c . c1 c . c2 Sin i Sin r = n1 sin i = n2 sin r

7. Total internal reflection The inside surface of water or a glass block can act like a mirror.

8. 1 Internal reflection A light ray hits the inside face of a semicircular block as follows. air glass What will happen?

9. 1 Internal reflection For a small angle of incidence • The incident ray splits into 2 rays. refracted ray air glass reflected ray incident ray

10. 1 Internal reflection As you increase the angle of incidence: The angle of refraction increases refracted ray Until the angle of refraction = 90o This angle of incidence is called the critical angle. air glass reflected ray incident ray

11. 1 Internal reflection As you increase the angle of incidence: The angle of refraction increases Until the angle of refraction = 90o refracted ray This angle of incidence is called the critical angle. As the angle of incidence increases even more, there will be no refracted ray. ALL will be reflected air glass reflected ray incident ray

12. 1 Internal reflection This angle of incidence is called the critical angle. air glass c c reflected ray incident ray • This is called total internal reflection.

13. Critical angle and refractive index C C refracted ray air glass incident ray reflected ray 1 sin i 1 sin C = = n n sin 90º 1 sin r 1 1 ( ) C = sin1  n = or n sin C

14. Critical angle and refractive index Refractive index Critical angle Medium • critical angles of different materials Glass 1.50–1.70 30–42 49 Water 1.33 Perspex 1.5 42 Diamond 2.42 24

15. 45 45 • If light rays strike the inside face at an angle > 42, glass prism behaves like a perfect mirror. 45 45 45 45

16. Example 1 E 30 O 18 A ray of light travelling in the direction EO in air enters a rectangular block. angle of incidence = 30 angle of refraction = 18 (a) Find the refractive index n of the block. (b) Find the critical angle C of the ray for the block.

17. Example 1 sin 30 sin 18  n = =1.62 1 1.62 (b) C = sin1 = sin1 1 n (a) By Snell’s law, n sin 18 = 1  sin 30 = 38.1

18. Example 1 C D 30 B A (c) If the ray is incident on surface BC, from which surface and at what angle will the ray leave the block?

19. Example 1 C D 30 32.3 57.7 32.3 B A (c) The ray comes out from surface AD. The outgoing angle from normal is 60. 60

20. Q1 Which of the following… A B C D Which of the following angles is the critical angle of glass?

21. Q2 A horizontal light ray is… A B 45 C A horizontal light ray hits a prism as shown. What happens to the light ray?

22. Q3 If the breaker is 12 cm tall... A water small stone If the beaker is 12 cm tall and 8 cm wide, Can we always see the small stone below water from side A? (Given: Refractive index of water = 1.33)

23. Q3 If the breaker is 12 cm tall... A water small stone Critical angle of water 48.8 0.752 1.33

24. Q3 If the breaker is 12 cm tall... A water small stone For the light ray from the stone reaching side A, Maximum angle of incidence on side A 12 56.3 8

25. Q3 If the breaker is 12 cm tall... A water small stone greater Since is ________ than the critical angle, ______________________ occurs and we _________ always see the stone form side A. total internal reflection cannot

26. Refraction at a boundary between two transparent substances Summary n1 sin i = n2 sin r n for air = 1 If the light ray is going fromairto another substance, then n1 = 1 r i sin i = n2 sin r n2 = sin i sin r If the light ray is going fromthe other substance to air, then n2 = 1 n2 n1 n1 sin i = sin r n1 = sin r sin i If total internal reflection happens then i = ic and r = 90 and sin r = 1 n1 sin ic = n2 sin ic = n2 Since sin any angle is < 1 n2 must be smaller than n1 n1

27. Summary questions page 195 • 1. • state two conditions for a light ray to undergo total internal reflection at a boundary between two transparent substances. • Calculate the critical angle for i. glass of refractive index 1.52 and air, ii. water of refractive index 1.33 and air. • Ans • Conditions: • n1>n2 • i > ic • b. i. glass of refractive index 1.52 and air • sin ic = 1/1.52 ic = 41.1º • ii. water of refractive index 1.33 and air. • sin ic = 1/1.33 ic = 48.8º

28. Summary questions page 195 2. a. show that the critical angle at a boundary between glass of refractive index 1.52 and water of refractive index 1.33 is 61º ic = 61º sin ic = n2 sin ic = 1.33 n1 1.52 b. The figure shows the path of a light ray in water of refractive index 1.33 directed at an angle of incidence of 40º at a thick glass plate of refractive index of 1.52 Calculate: i. the angle of refraction of light ray at P ii. The angle of incidence of the light ray at Q air i. n1 sin i = n2 sin r glass water 1.33 sin 40 = 1.52 sin r Q P sin r = 1.33 sin 40 / 1.52 40º r = 34º ii. Angle at Q = 34º

29. 3. A window pane made of glass of refractive index 1.55 is covered on one side only with a transparent film of refractive index 1.40. • Calculate the critical angle of the film-glass boundary • A light ray in air is directed at the film at an angle of incidence 45º as shown in the diagram. Calculate • the angle of refraction in the film. • The angle of refraction of the ray where it leaves the pane. sin ic = 1.40 sin ic = n2 ic = 64.6º 1.55 n1 i. n1 sin i = n2 sin r film air glass air 1x sin 45 = 1.4 sin r sin r = 1x sin 45 / 1.4 45º r = 30.3º ii. Next slide

30. 3. A window pane made of glass of refractive index 1.55 is covered on one side only with a transparent film of refractive index 1.40. • Calculate the critical angle of the film-glass boundary • A light ray in air is directed at the film at an angle of incidence 45º as shown in the diagram. Calculate • the angle of refraction in the film. • The angle of refraction of the ray where it leaves the pane. ii. From film to glass n1 sin i = n2 sin r film air glass air 1.44 sin 30.3 = 1.55 sin r r = 28º from glass to air n1 sin i = n2 sin r 45º 28º 1.55 sin 28 = 1x sin r 30.3º r = 46.6º

31. 4. • In a medical endoscope, the fibre bundle used to view the image is coherent. • What is meant by coherent • Explain why this fibre bundle needs to be coherent • b. • Why is an optical fibre used in communication composed of a core surrounded by a layer of cladding of lower refractive index? • Why is it necessary for the core of an optical communications fibre to be narrow? The fibre ends at each end are at the same relative positions So the image is not distorted Since sin ic = n2 Total internal reflection happens when n2<n1 n1 If the core was wide, some light rays might travel along its axis and take a shorter route than the light ray that is being internally reflected.