Snell’s Law
•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).
Angle of
incidence (i)
Angle of
refraction (r)
sin (i)/sin(r)
i
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.
I1 and r1 are the angles of incidence and refraction at
the point where the light ray enters the glass block.
The refractive index of the block, n = Sin i1
Sin r1
i1
r1 i
2
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
n=
Sin i1
Sin r1
=
Sin r2
Sin i2
or
1 .
n
=
Sin i2
Sin r2
This will apply whenever the light
ray goes from a substance to air
r2
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
1 .
n
Sin i
Sin r
=
i
glass
1 .
1.5
1 .
1.5
=
=
Sin 30
Sin r
Sin 30
Sin r
Sin r = 1.5 x sin30
r = 48.6º
Refraction by a prism
Q. A light ray enters an
equilateral triangular prism of
refractive index 1.55 at an angle
of incidence of 35º .
i1
r1
i2
r2
a. Calculate the angle of
refraction in the glass.
1.55 =
Sin 35
Sin r
Sin r1 = Sin 35
1.55
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
1 .
n
=
Sin i2
Sin r2
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.
Where: c is the speed of light in vacuum or air (3x108ms-1)
n= c
cs
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
n= λxf
λc x f
n= λ
λc
=
λ
x
f
As the wave speed decreases, the wavelength
decreases and the frequency stays the same.
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
Sin i
Sin r
=
c1.
c2
i
This can be rearranged as:
1.
Sin i
c1
n1
=
1 . Sin r
c2
=
c . Sin r
c2
Multiplying both sides x c
c.
Sin i
c1
n1 sin i = n2 sin r
n2
Total internal reflection
The inside surface of water or a glass block
can act like a mirror.
1
Internal reflection
A light ray hits the inside face of a
semicircular block as follows.
air
glass
What will happen?
1
Internal reflection
For a small angle of incidence
• The incident ray splits into 2 rays.
air
glass
incident ray
1
Internal reflection
As you increase the angle of incidence:
The angle of refraction increases
Until the angle of refraction = 90o
This angle of incidence is called the critical angle.
air
glass
incident ray
1
Internal reflection
As you increase the angle of incidence:
The angle of refraction increases
Until 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
air
glass
incident ray
1
Internal reflection
This angle of incidence is called the critical angle.
air
glass
incident ray
c
c
reflected ray
• This is called total internal reflection.
Critical angle and refractive index
air
glass
refracted ray
C C
incident ray
reflected ray
1
sin i
=
n
sin r
1
n=
sin C
1
sin C
=
n
sin 90º 1
or
C=
sin1
(1)
n
Critical angle and refractive index
• critical angles of different
materials
Refractive
Critical
Medium
index
angle
Glass
1.50–1.70
30–42
Water
1.33
49
Perspex
1.5
42
Diamond
2.42
24
• If light rays strike the inside face at an
angle > 42, glass prism behaves like
a perfect mirror.
45
45
45
45
45
45
Example 1
A ray of light travelling in the direction EO in
air enters a rectangular block.
E
angle of incidence = 30
30
angle of refraction = 18
O
18
(a) Find the refractive index n of the block.
(b) Find the critical angle C of the ray
for the block.
Example 1
(a) By Snell’s law, n sin 18 = 1  sin 30

(b) C =
sin 30
n=
=1.62
sin 18
sin1
1 = sin1 1 = 38.1
1.62
n
Example 1
(c)
If the ray is incident on surface BC,
from which surface and at what angle
will the ray leave the block?
D
C
30
A
B
Example 1
(c)
The ray comes out from surface AD.
The outgoing angle from normal is 60.
C
D
60
A
32.3
57.7
32.3
30
B
Q1 Which of the following…
Which of the following angles is the critical
angle of glass?
A
B
C
D
Q2 A horizontal light ray is…
A horizontal light ray hits a prism
as shown.
What happens to the light ray?
A
B
C
45
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
water A
stone below water from side A?
small stone
(Given: Refractive index of water = 1.33)
Q3 If the breaker is 12 cm tall...
Critical angle of
1
1 
 sin 
 ( 1.33
water

48.8
  sin1 ( 0.752 )  ________
)
water A
small stone
Q3 If the breaker is 12 cm tall...
For the light ray from the stone reaching side
A,
Maximum angle of incidence  on side A
height

1 
 tan 

 width of the base 
water A
12  
1  
56.3
  _________
 tan 
  8 
small stone
Q3 If the breaker is 12 cm tall...
greater than the critical angle,
Since is ________
total internal reflection occurs and we
______________________
cannot always see the stone form side A.
_________
water A
small stone
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 from air to
another substance, then n1 = 1
sin i = n2 sin r
n2 = sin i
sin r
If the light ray is going from the
other substance to air, then n2 = 1
r
i
n1
n2
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
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
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
i.
n1 sin i = n2 sin r
water glass air
1.33 sin 40 = 1.52 sin r
Q
P
sin r = 1.33 sin 40 / 1.52
r = 34º
ii. Angle at Q = 34º
40º
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
sin ic = 1.40
sin ic = n2
ic = 64.6º
1.55
n1
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.
i.
n1 sin i = n2 sin r
air
1x sin 45 = 1.4 sin r
sin r = 1x sin 45 / 1.4
ii.
r = 30.3º
Next slide
45º
film
glass
air
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.
ii. From film to glass n1 sin i = n2 sin r
air
film
glass
1.44 sin 30.3 = 1.55 sin r
r = 28º
from glass to air n1 sin i = n2 sin r
1.55 sin 28 = 1x sin r
r = 46.6º
45º
30.3º
28º
air
4.
a. In a medical endoscope, the fibre bundle used to view the image is
coherent.
i. What is meant by coherent
The fibre ends at each end are at the same relative positions
ii. Explain why this fibre bundle needs to be coherent
So the image is not distorted
b.
i. Why is an optical fibre used in communication composed of a core
surrounded by a layer of cladding of lower refractive index?
Since sin ic = n2
n1
Total internal reflection happens when n2<n1
ii. Why is it necessary for the core of an optical communications fibre to
be narrow?
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.