1. Good AfternoonACTIVITY

2. Refraction
of
Light

3. What is Refraction?
 Refraction is the bending of light as it
travels from one medium to another.

4. Refraction
 In vacuum, light travels with a speed
c = 3.00 x 108 m/s
 In a transparent medium, (ex. water or
glass) the speed of light is slower than
its speed in vacuum.
 WHY? Ans. interactions between photons
and molecules of the medium

5. Refraction
What EXACTLY is light doing when it reaches a
new medium?
Air
Water
Incident ray Reflected ray
Refracted ray
Angle of
Incidence
Angle of
Reflection
Angle of
Refraction

6. Refraction
Air
Water
Normal
Incident ray
Refracted ray
θi
θr
Normal
is a line drawn
perpendicular to the surface
Incident ray
is the light ray in 1st
medium (air)
Refracted ray
is the light ray in the
2nd medium (water)
Angle of incidence, i
is the angle between the
incident ray & the normal
Angle of refraction, r
is the angle between the
refracted ray & the
normal
Law of Refraction #1
 The incident ray, refracted ray, and normal lie in one play.

7. Air
Glass
Normal
Higher speed (lower index of refraction)
Lower speed (higher index of refraction)
Refraction
θi
θr
Refracted ray bends
away from the normal
Original path of
light ray

8. Refraction
Air
Glass
Higher speed (lower index of refraction)
Lower speed (higher index of refraction)
Refracted ray
bends towards the
normal
θi
Normal
θr Original path
of a light ray

9. Refraction
Air
(a) Less dense to denser medium (b) Denser to less dense medium
Air
Glass
Glass
θi
θr
θi
θr
(a) When light ray enters a medium with smaller index of refraction
to a medium with larger index of refraction, it bends TOWARDS the
normal.
(b) When light ray enters a medium with larger index of refraction, it
bends AWAY the normal.
Law of refraction #2
 When a ray of light enters a denser medium, it bends TOWARDS the
normal
 When a ray of light enters a less dense medium, it bends AWAY from
the normal.

10. What is “index of refraction”?
The index of refraction (n) for a material is
the ratio of the speed of light (c) in vacuum to
that of the speed (v) in the medium.
The ratio of the two speeds can be compared
c
n
v

c
v
c
n
v

Medium Refracted Index (n)
Vacuum 1.00
Air 1.0003
Water 1.33
Ethanol 1.36
Glycerine 1.47
Crown Glass 1.52
Quartz 1.54
Flint Glass 1.61
Diamond 2.42

11. Sample Problem
What is the speed of light in a diamond if
its index of refraction is 2.42?
Light travels from air (n = 1) into glass, where its velocity reduces to
only 2 x 108 m/s.
What is the index of refraction for glass?
vair = c
vG = 2 x 108 m/s
Glass
Air
8
8
3 x 10 m/s
2 x 10 m/s
c
n
v
 
Solution:
Answer:
n= 1.50
What is the speed of light in quarts (1.54)?
Solution
V= c/n
V=
3.00𝑥108 𝑚/𝑠
1.54
Given
n= 1.54
c= 3.00× 𝟏𝟎 𝟖 m/s
v= ?
Answer
V= 1.95x𝟏𝟎 𝟖
m/s

12. Law of refraction
1. The incident ray, refracted ray, and normal lie in one plane.
2. When a ray of light enters a denser medium, it bends
TOWARDS the normal.
When a ray of light enters a less dense medium,
it bends AWAY from the normal.
3. Snell’s Law is the relationship between the angle of
incidence and angle of refraction and the angle of refraction of
both materials.
How can we figure out the angle of our
refracted ray?
We use Snell’s Law
Willebrord van Roijen Snell
(1580-1626)
Sample Problem
A ray of light traveling through air is
incident on a smooth surface of water
at an angle of 30° to the normal.
Calculate the angle of refraction for the
ray as it enters the water.
Air
Water
𝒏 𝟏sin𝐬𝐢𝐧 𝜽 𝒊 = 𝒏 𝟐 𝐬𝐢𝐧 𝜽 𝒓
Formula
𝐬𝐢𝐧 𝜽 𝒓
𝒏 𝟏 𝐬𝐢𝐧 𝜽 𝒊
𝒏 𝟐
Given
𝒏 𝟏= 1.0003 (air)
𝒏 𝟐= 1.33 (water)
sin 𝜽 𝒊 = 𝟑𝟎°
Find 𝜽𝒓 = ?
Solution
sin 𝜽 𝒓 =
𝟏. 𝟎𝟎𝟎𝟑 sin 𝟑𝟎°
𝟏. 𝟑𝟑
𝐬𝐢𝐧 𝜽𝒓 = 𝟎. 𝟑𝟕𝟔𝟎𝟓𝟐𝟔𝟑𝟏𝛉𝐫 = 𝐬𝐢𝐧−𝟏
𝟎. 𝟑𝟕𝟔𝟎𝟓𝟐𝟔𝟑𝟏
Answer:
𝜽𝒓 = 𝟐𝟐. 𝟎𝟗°

13. Mmmm..…….
Fish and
Chips!!!
Normal
Effects of Refraction
Real depth
Apparent
depth
To an observer standing at
the side of a swimming
pool, objects under the
water appear to be nearer
the surface than they really
are.
It can be shown mathematically that index of refraction is the ratio
of real depth to apparent depth.
Formula:
n=
𝑹𝒆𝒂𝒍 𝒅𝒆𝒑𝒕𝒉
𝑨𝒑𝒑𝒂𝒓𝒆𝒏𝒕 𝒅𝒆𝒑𝒕𝒉
Sample Problem
A grizzley bear is seating on a rock in the middle of a calm river,
when she observes a fish directly below. If the apparent deep of a fis
is 0.40m, what is the actual depth of a fish?
Given
n= 1.33 (water)
App. Depth= 0.40m
Find Actual depth=??
Solution
Real Depth= n* apparent depth
Real depth= (1.33)(0.40m)
Real depth= 0.53m

14. Total Internal Reflection
When light passes at an angle from a medium of higher index to
one of lower index, the emerging ray bends away from the normal.
Water
Air
light
i = r
Critical
angle
qc
900
When the angle reaches a certain
maximum, it will be reflected
internally.
The critical angle is defined
as the angle of incidence in
the optically denser medium
for which the angle of
refraction in the optically less
dense medium is 90°.
n
So the question is , how can
you calculate the critical
angle?
Remember, it is when the
refracted ray is equal to 90
degrees
ɵc = sin-1 (𝒏 𝟐/𝒏 𝟏)
Sample Problem
What is the critical angle for
water-air interface
Given
n1= (water) 1.33
n2= (air) 1.0003
𝜽𝒄 = sin−𝟏
𝟏. 𝟎𝟎𝟎𝟑
𝟏. 𝟑𝟑
𝜽𝒄 = 𝟒𝟖. 𝟕𝟕°