•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.

i

r

Angle of incidence (i)

Angle of refraction (r)

   

   

   

   

   

   

   

   

   

sin (i)/sin(r)

 

 

 

 

 

 

 

 

 

Snell’s Law

The refractive index of the block, n = Sin i1Sin r1

i1

r1

I1 and r1 are the angles of incidence and refraction at the point where the light ray enters the glass block.

i2

r2

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

Sin i1Sin r1

n = = Sin r2

Sin i2

or

Sin i2Sin r2

1 . n =

This will apply whenever the light ray goes from a substance to air

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

i

glass

Sin iSin r

1 . n =

Sin 30 Sin r

1 . 1.5 =

Sin 30 Sin r

1 . 1.5 =

Sin r = 1.5 x sin30

r = 48.6º

Refraction by a prism

1.55 = Sin 35Sin r

i1 r1i2

r2Q. 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.

Sin r1 = Sin 351.55 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 i2Sin r2

1 . n = Sin i2

Sin r2

1 . n = Sin r2 = 0.96 r2 = 73.8

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.

n = ccs

Where: c is the speed of light in vacuum or air (3x108ms-1)

cs is the speed of light in the substance

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

n = λ x fλc x f

As the wave speed decreases, the wavelength decreases and the frequency stays the same.

n = λ λc

Refraction at a boundary between two transparent substances

i

r

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:

Sin i Sin r =

c1.

c2

This can be rearranged as:

n1 sin i = n2 sin r

n1 n2

1 .

c2

1 .

c1

Sin i = Sin r

c .

c2

c .

c1

Sin i = Sin r

Multiplying both sides x c

The inside surface of water or a glass block can act like a mirror.

Total internal reflection

1 Internal reflection

A light ray hits the inside face of a semicircular block as follows.

What will happen?

airglass

1 Internal reflection

• The incident ray splits into 2 rays.

incident ray

reflected rayrefra

cted ray

airglass

For a small angle of incidence

1 Internal reflection

incident ray

reflected rayrefra

cted ray

airglass

As you increase the angle of incidence:

The angle of refraction increases

Until the angle of refraction = 90oThis angle of incidence is called the critical

angle.

1 Internal reflection

incident ray

reflected rayrefra

cted ray

airglass

As you increase the angle of incidence:

The angle of refraction increasesUntil the angle of refraction = 90o

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

1 Internal reflection

incident ray reflected ray

airglass

This angle of incidence is called the critical angle.

c c

• This is called total internal reflection.

Critical angle and refractive index

incident ray reflected ray

airglass

C C

refracted ray

sin C =

1 sin C

n = C = sin1or 1 n

( )

sin 90º1n sin r

= sin i1n 1

Critical angle and refractive index

• critical angles of different materials Medium

Refractive index

Critical angle

1.50–1.70 30–42Glass

Water

Perspex

Diamond

1.33 49

1.5 42

2.42 24

4545

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

4545

4545

Example 1A ray of light travelling in the direction EO in air enters a rectangular block.

angle of incidence = 30angle of refraction = 18

(a) Find the refractive index n of the block.

(b) Find the critical angle C of the ray for the block.

30

18

E

O

Example 1(a) By Snell’s law, n sin 18 = 1 sin 30

n = =1.62sin 30sin 18

(b) C = sin11n

= sin1 11.62 = 38.1

Example 1(c) If the ray is incident on surface BC,

from which surface and at what angle will the ray leave the block?

30

A B

CD

30

A B

CD

Example 1The ray comes out from surface

AD. The outgoing angle from normal is 60.

32.357.7

32.3

60

(c)

Q1 Which of the following…Which of the following angles is the critical angle of glass?

A B

DC

Q2 A horizontal light ray is…

A horizontal light ray hits a prism as shown.

What happens to the light ray?45

A B

C

water A

small stone

Q3 If the breaker is 12 cm tall...

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)

Q3 If the breaker is 12 cm tall...

Critical angle of water

________)(sin)(

1sin 11

water A

small stone

1.330.752 48.8

Q3 If the breaker is 12 cm tall...For the light ray from the stone reaching side A,

water A

small stone8

12 56.3

Maximum angle of incidence on side A

base the of widthheight

tan1

_________tan1

Q3 If the breaker is 12 cm tall...

Since is ________ than the critical angle,______________________ occurs and we _________ always see the stone form side A.

water A

small stone

total internal reflectiongreater

cannot

Refraction at a boundary between two transparent substances

Summary

i

r

n1 sin i = n2 sin r

n1 n2

n for air = 1

If the light ray is going from air to another substance, then n1 = 1

sin i = n2 sin r n2 = sin isin r

If the light ray is going from the other substance to air, then n2 = 1

n1 sin i = sin r n1 = sin rsin i

If total internal reflection happens then i = ic and r = 90 and sin r = 1

n1 sin ic = n2 sin ic = n2

n1

Since sin any angle is < 1

n2 must be smaller than n1

Summary questions page 195

1.

a. state two conditions for a light ray to undergo total internal reflection at a boundary between two transparent substances.

b. Calculate the critical angle for i. glass of refractive index 1.52 and air, ii. water of refractive index 1.33 and air.

Ans

a. 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

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

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

glass airwater

40º

sin ic = n2

n1

sin ic = 1.331.52

ic = 61º

n1 sin i = n2 sin ri.

1.33 sin 40 = 1.52 sin r

sin r = 1.33 sin 40 / 1.52

r = 34º

ii. Angle at Q = 34º

PQ

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.

a. Calculate the critical angle of the film-glass boundary

b. A light ray in air is directed at the film at an angle of incidence 45º as shown in the diagram. Calculate

i. the angle of refraction in the film.

ii. The angle of refraction of the ray where it leaves the pane.

glass airfilm

45º

sin ic = n2

n1

sin ic = 1.401.55

ic = 64.6º

i. n1 sin i = n2 sin r

1x sin 45 = 1.4 sin rair

sin r = 1x sin 45 / 1.4

r = 30.3º

ii. Next slide

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.

a. Calculate the critical angle of the film-glass boundary

b. A light ray in air is directed at the film at an angle of incidence 45º as shown in the diagram. Calculate

i. the angle of refraction in the film.

ii. The angle of refraction of the ray where it leaves the pane.

glass airfilm

45º

airii. From film to glass n1 sin i = n2 sin r

1.44 sin 30.3 = 1.55 sin r

r = 28º

30.3º 28º

from glass to air n1 sin i = n2 sin r

1.55 sin 28 = 1x sin r

r = 46.6º

4.

a. In a medical endoscope, the fibre bundle used to view the image is coherent.

i. What is meant by coherent

ii. Explain why this fibre bundle needs to be coherent

b.

i. Why is an optical fibre used in communication composed of a core surrounded by a layer of cladding of lower refractive index?

ii. 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

n1

Total internal reflection happens when n2<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.