Chapter 4: Refraction
Refraction is the bending of a wave as it passes from one medium to another.
When light travels from a rarer to a denser medium it is refracted towards the normal and vice versa.
The Laws of Refraction of Light
1. The incident ray, the normal and the refracted ray all lie on the same plane.
2.
Sini
 constant
Sinr
This constant is called the refractive index between the two media, and is given the symbol Ŋ (big fancy n).
The second law of refraction is also known as Snell’s Law*.
In words:
Snell’s Law states that the ratio of the sin of the angle of incidence to the sin of the angle of refraction when light
travels from one medium to another is a constant.
This leads to the following definition:
The Refractive Index of a Medium is the ratio of the sine of the angle of incidence to the sine of the angle of
refraction when light travels from a vacuum into that medium*.
In other words if you see the phrase “the Refractive Index of glass is 1.5”, it means that when light travels from air
into glass the refractive Index is 1.5*.
The Refractive Index between two Media is the ratio of the sine of the angle of incidence to the sine of the angle
of refraction when light travels from one of those media into the other.
n
The refractive index between water and glass is written as w g.
For any two media x and y:
x y 
1
y x
So for example if you’re told that “the refractive index for water is 1.33”, it means when light is going from air to
n
water, i.e. a g = 1.33.
n
This in turn means that if light is going from water to air, the refractive index is 1/1.33 or g a = 1/1.33 = 0.75
See problems 1 – 4 page 30, then try question 1 – 8 page 31.
Real Depth and Apparent Depth
Re alDepth

ApparentDepth
See problems 5 and 6, page 34, and try questions 1 – 5 page 34.
Question 5 is challenging but worth a go – draw a diagram.
Refractive Index in Terms of Relative Speeds
C1

C2
Where C1 is the speed of light in the first medium and C2 is the speed of light in the second medium*.
Usually one of these media will be air.
Now look at problems 7 and 8, then try questions 1–4, page 36.
Total Internal Reflection
When the angle of incidence in the denser of two media reaches a certain (critical) angle, the light no longer refracts
outward, but instead travels along the surface (see diagram).
The angle of refraction is now 900, and this leads to a definition of the critical angle.
When light travels from a denser to a rarer medium, the angle of incidence which causes the angle of refraction
to be 900 is called The Critical Angle (for those two media).
To Demonstrate Total Internal Reflection*
Relationship between Critical Angle and Refractive Index
1
  SinC
Where C represents the critical angle*.
Remember Refractive Index of a medium is defined in relation to light going from air into that medium.
Now look over examples 9 and 10, page 38. Then try questions 1 – 6, page 38
Using a 450– 900 – 450 prism to turn a ray of light through
(i)
(ii)
900,
1800
Applications of Total Internal Reflection
1.
2.
3.
4.
Reflective road signs
Mirages
Prism reflectors
Optical Fibres
Optical Fibres
It consists of a glass pipe coated with a second material of lower refractive index. Light enters one end of the fibre at
an angle greater than the critical angle, resulting in Total Internal Reflection at the boundary between the two
materials. This gets repeated all along the glass pipe.
If the second cladding material wasn’t there or had a refractive index greater than that of the central core Total Internal
Reflection would not occur and the light would simply escape out.
Applications
1. Telecommunications
2. Medicine (endoscopes)
Advantages of optical fibres over copper conductors
Less interference / boosted less often / cheaper raw material / occupy less space / more information (carried) in the
same space / flexible for inaccessible places/ do not corrode
Summary of Formula
Sini

Sinr
1
Re alDepth

ApparentDepth
C1

C2
x y 
  SinC
1
y x
Mandatory Experiment:
To verify Snell’s Law and use it to measure the refractive index of a solid.
Extra Credit
*When light travels from a vacuum into that medium
In practice we consider air and a vacuum to be equivalent.
*In other words if you see the phrase “the Refractive Index of glass is 1.5”, it means that when light travels
from air into glass the refractive Index is 1.5
This is important to remember because if light is going in the other direction (i.e. from glass to air, the refractive index
will not be 1.5.
This often causes confusion, especially when we go on to define the critical angle.
*Ŋ = c1/c2
Why is "c" used for the speed of light?
From the Latin 'celeritas', meaning speed, as an alternative to v (for particles).
*To Demonstrate Total Internal Reflection
What the text book doesn’t mention, but what you will notice when you carry out this demonstration, is that because
the ray of light is actually quite wide, the inner part of the ray will be totally internally reflected while the outer part
will get refracted outwards. You will therefore actually end up with two rays.
*
1
  SinC
The derivation of this formula is on pages 37/38 – you don’t need to know it, but it might be useful to see if you can
follow it.
Experiments:
There are three mandatory experiments on the syllabus to do with refraction
1. To verify Snell’s Law
2. To measure the refractive index of a liquid or a solid.
The or in italics implies that you must be given the option, and therefore the exam question cannot specify measuring
the refractive index of a liquid.
But in verifying Snell’s Law, we plot a graph of Sin i against Sin r, and show that because the graph is a straight line
going through the origin, the two variables are directly proportional.
To calculate the refractive index we then simply calculate the slope of the graph, and if Sin i is on the y-axis, the slope
corresponds to the refractive index.
No need to worry about measuring the refractive index of any messy liquids!
Hints for carrying out the experiment:
Start with a large angle for incidence and measure the corresponding angle of refraction.
Now reduce this angle at roughly regular intervals each time to ensure a wide range of values.
A common mistake is to measure the angle at the wrong side of the normal; it must be the angle between the normal
and the incident or refracted ray.
Demonstrations
Coke can in a large beaker of water; the size of the can becomes magnified due to refraction.
Pencil bending in a beaker of water,
Place a penny in a basin of water and gradually lower your head until you can no longer see the penny due to the top
of the basin. Now add water (or get someone else to do it for you). The penny magically reappears!
Large Toy-Car and Sandpaper; car changes direction when it passes from smooth surface to sandpaper and reverts to
its original direction when it leaves. This is because the two front wheels are travelling at different speeds when they
are on different media.
Swell-Gel hanging in a net ‘disappears’ in a beaker of water.
Glass beakers and/or stirrers ‘disappear’ in a beaker of baby-oil.
Both of the previous demos are due to almost identical refractive indices of the different materials.
Exam Questions and Answers
2007 No. 5 (e)
The refractive index of a liquid is 1.35, what is the critical angle of the liquid?
ng =1/ sin c
c = 47.8 (-1 penalty for answer 53.1o …. gradian mode used)
2006 Question 7 {first part}
What is meant by the refraction of light? (6)
The bending (of light)
on passing from one medium to another
2005 Question 3
In an experiment to verify Snell’s law, a student measured the angle of incidence i and the angle of refraction r for a
ray of light entering a substance. This was repeated for different values of the angle of incidence. The following data
was recorded.
i/degrees
20
30
40
50
60
70
r/degrees
14
19
26
30
36
40
Describe, with the aid of a diagram, how the student obtained the angle of refraction. (9)
pins / ray box (to obtain incident and refracted rays)
diagram to show: outline of block, incident and refracted ray, normal
measure angle between refracted ray and normal (using a protractor / trig.)
Draw a suitable graph on graph paper and explain how your graph verifies Snell’s law. (18)
sin i
sin r
0.34
0.24
0.50
0.33
0.64
0.44
0.77
0.50
0.87
0.59
0.94
0.64
Sin i and sin r correct values ( -1 for each incorrect value)
Labelled axes
At least 5 points plotted correctly
Straight line drawn
Good distribution
Conclusion e.g. sin i proportional to sin r / straight line through the origin.
From your graph, calculate the refractive index of the substance.
Correct method for slope e.g. ( m = ) y2 – y1 / x2 – x1
Substitute coordinates of two points on the graph
n = 1.49 (accept range: 1.44 – 1.50)
o
The smallest angle of incidence chosen was 20 . Why would smaller values lead to a less accurate result? (4)
Greater percentage error (in these readings)
2004 Question 12(b)
Give two reasons why the telecommunications industry uses optical fibres instead of copper conductors to
transmit signals. (6)
Less interference / boosted less often / cheaper raw material / occupy less space / more information (carried) in the
same space / flexible for inaccessible places/ do not corrode
Explain how a signal is transmitted along an optical fibre.
Light ray/signal introduced at one end of fibre; … (stated or implied) 3
Strikes interface/boundary at i > ic 3
Total internal reflection occurs 3
( -1 if no reference to light)
An optical fibre has an outer less dense layer of glass. What is the role of this layer of glass?
Signal/light (then always) remains in denser medium / t.i.r. occurs / prevents light passing from one fibre to an
adjacent one / prevents light escaping / prevents damage to surface of the core, etc.
An optical fibre is manufactured using glass of refractive index of 1.5.
Calculate the speed of light travelling through the optical fibre. (9)
ng = c1/c2 or ca /cg
1.5 = 3 x 108/ vg
vg = 2.0 x 108 (ms-1)
2003 Question 5 (f)
Calculate the critical angle for diamond. The refractive index of diamond is 2.4.
1/Sin C
24.60