Introduction
 Refraction refers to the bending of light when it passes from one medium into another of different optical density.
 This is because as light passes through different media its velocity changes.
 The bending occurs at the boundary or interface of the two media.
 The refracted ray may bend away or towards the normal depending on the optical density of the second medium with respect to the first medium.
 Generally, a ray passing from an optically denser medium into a less optically dense (rarer) medium is bent away from the normal after refraction.
 If the ray passes from a rarer medium into an optically denser medium then it is bent towards the normal.
 It is easier to tell which medium is optically denser by simply comparing the angle between the incident ray and the normal and that between the refracted ray and the normal.
 The medium with a smaller angle (of incidence or refraction) is the optically denser medium.
 However, when the ray strikes the interface perpendicularly (normally) it passes undeviated (without bending). This is because the angle of incidence is zero.
 In figure (b) above, only the direction of the light has been reversed leaving the angles the same. However, i now becomes r while r becomes i. The principle that makes it possible to reverse the direction of light keeping the sizes of the angles the rays make with the normal the same is called the principle of reversibility of light.
 The study of refraction of light helps us understand the following common phenomena:
 Why a stick appears bent when part of it is in water.
 Why a coin at the base of a beaker of water appears nearer the surface than it actually is.
 Why the stars twinkle.
 Why the sun can still be seen sometimes before it rises or even after setting.
 Why the summer sky appears blue.
 The formation of the rainbow.
Refraction in Glass
This can be investigated by the following steps:
 Fix a white plain paper on a soft board using drawing pins. Place the glass block with its larger surface on the plain paper and trace its outline.
 Remove the glass block and then draw a normal through point O. Draw a line making an angle say i = 30^{o} with the normal as shown above.
 Replace the glass block onto the outline and stick two pins P_{1} and P_{2} along the line such that they are upright and about 6cm apart.
 From the opposite side of the block, view the two pins and stick two pins P_{3} and P_{4} such that the four pins appear on a straight line. Join the positions of P_{3} and P_{4} using a straight line and produce the line to meet the outline at O’.
 Draw another normal at O’ and then join O to O’. Measure angles r and e .
 Repeat the above steps for other values of i=40^{o}, 50^{o} and 60^{o} . Complete the table below:
Angle of incidence, i^{o} 30^{o} 40^{o} 50^{o} 60^{o} Angle of refraction, r^{o} e^{o} Sin i Sin r ^{Sin} ^{i}/_{sin r}  Plot a graph of Sin i against Sin r. determine the slope of your graph.
Observations
 The ratio of Sin i to Sin r is a constant.
 The graph of Sin i against Sin r is a straight line through the origin.
 The slope of the graph is equal to the ratio of Sin i to Sin r in the table.
The Laws of Refraction and Refractive Index
There are two laws of refraction:
 The incident ray, refracted ray and the normal at the point of incidence all lie in the same plane.

Snell’s law : it states that the ratio of sine of angle of incidence to the sine of angle of refraction is a constant for a given pair of media.
i.e. ^{Sin i}/_{Sin r} = a constant.
The constant is referred to as the refractive index, η of the second medium with respect to the first medium.
The first medium is that medium in which the incident ray is found while the second medium is that medium where the refracted ray is found. It is denoted as _{1}η_{2}.
Hence in the previous section, the ratio ^{Sin i}/_{Sin r} is the refractive index of glass with respect to the air since the light passed from air into glass block.
However, when light passes from vacuum into another medium, it is referred to as absolute refractive index.
Therefore for absolute refractive index, the angle of incidence i is found in a vacuum.
i.e. absolute refractive index= sin i(in vacuum)/sin r(in the second medium).
Recall:
_{1}η_{2} =^{sin i}/_{sin r}By the principle of reversibility of light, r now becomes i and i becomes r i.e. the incident ray is now found in the second medium.
Hence _{2}η_{1} =^{sin r}/_{sin i}But ^{sin r}/_{sin i}=^{1}/_{(sin i/sin r)}=^{1}/_{1η2}Therefore _{2}η_{1} =^{1}/_{1}η_{2}.
The table below shows some materials and their refractive indices:Material Refractive index Ice 1.31 Crown Glass 1.50 Water 1.33 Alcohol 1.36 Kerosene 1.44 Diamond 8.42
Example 8.1
 In the figure below, calculate the angle of refraction r given that the refractive index of the glass is 1.50.
By the principle of reversibility of light;
^{sin r}/_{sin 30o}= 1.50
sin r = 1.50 × sin 30^{o}r = sin^{1}(1.50 × sin 30^{o})= 48.6^{o}  A ray of light is incident on a flat glass surface as shown below:
Given that the refractive index of glass is 1.50, determine the angle of refraction for the ray of light.
1.50 = ^{sin 35}/_{sin r}
Sin r = ^{sin 35}/_{1.50}r=sin^{1}(^{sin 35}/_{1.50}) = 3.48^{o}
Refraction through Successive Media
 Consider a ray of light passing through a series of media as shown below:
 Suppose the boundaries are parallel, then:
_{a}η_{1} =^{sin i}/_{sin r1}………………………………. (i)
_{1}η_{2} =^{sin r1}/_{sin r2} …………………………… (ii)
_{2}η_{a} =^{sin r2}/_{sin i} ……………………………… (iii)  By the principle of reversibility of light;
_{a}η_{2} =^{sin i}/_{sin r2}……………………………. (iv)  Also, multiplying equations (i) and (ii), we get:
_{a}η_{1} × _{1}η_{2} = ^{sin i}/_{sin r1 }× ^{sin r1}/_{sin r2} = ^{sin i}/_{sin r2}.
Thus _{a}η_{2} = _{a}η_{1} × _{1}η_{2}.  Generally, _{1}η_{k} = _{1}η_{2} × _{2}η_{3} ×………….× _{k1}η_{k} .
Example 8.2
 A ray of light from air passes successively through parallel layers of water, oil, glass and then into air again. If the refractive indices of water, oil and glass are ^{4}/_{3}, ^{6}/_{5} and ^{3}/_{2} respectively and the angle of incidence in air is 60^{o} .
 Draw a diagram to show how the ray passes through the multiple layers.
 Calculate:
 The angle of refraction in water.
^{4}/_{3}= ^{sin 60}/_{sin r}r=sin^{1}(^{3sin60}/_{4})= 40.5^{o}  The angle of incidence at the oilglass interface.
_{o}η_{g} =^{sin q}/_{sin r}By the principle of reversibility of light, _{a}η_{g} =^{sin 60}/_{sin r} = ^{3}/_{2}
r= sin^{1}(^{2sin 60}/_{3}) =38.27^{o} .
Also, _{o}η_{g} = _{o}η_{a} × _{a}η_{g} =^{5}/_{4}Therefore, ^{5}/_{4}= ^{sin q}/_{sin 38.27}q=sin^{1}(^{5sin38.27}/_{4})=48.4^{o}
 The angle of refraction in water.
 Draw a diagram to show how the ray passes through the multiple layers.
Refractive Index in terms of Real and Apparent Depth
 This is on the basis that when an object at the base of a container filled with water is viewed perpendicularly it appears closer to the surface than it actually is. Consider the figure below:
 Hence, refractive index of water= Real depth/Apparent depth.
 When a graph of real depth against apparent depth is plotted, the graph obtained is a straight line through the origin and whose gradient is equal to the refractive index of the medium involved.
Example 8.3
 In a transparent liquid container, an air bubble appears to be 12cm when viewed from one side and 18cm when viewed from the other side. If the length of the tank is 40cm, where exactly is the air bubble?
Refractive index of glass= ^{(12+x)}/_{12} = ^{(18+2x)}/_{18}
x= ^{20}/_{5} = 4cm.
Therefore, the bubble is 3cm in the liquid from the lefthand side.  A microscope is focused on a mark on a horizontal surface. A rectangular glass block 30mm thick is placed on the mark. The microscope is then adjusted 2mm upwards to bring the mark back to focus. Determine the refractive index of the glass.
_{a}η_{g} =^{real depth}/_{apparent depth}= ^{30mm}/_{20mm }=1.50
Refractive Index in Terms of Velocity of Light
 Refraction occurs as a result of the different light velocity in different media. Basically, refractive index of any medium is the ratio of the velocity of light in a vacuum or air to the velocity of light in that medium; η m = velocity of light in vacuum/velocity of light in the medium.
Note that the velocity of light in a vacuum is 3.0 × 10^{8} m/s.
Generally, _{1}η_{2} =^{velocity of light in medium 1}/_{velocity of light in medium 2}.
Example 8.4
 The velocity of light in glass is 8.0 × 10^{8} m/s. Calculate:
 The refractive index of glass.
η_{g} = velocity of light in vacuum/velocity of light in glass= (3.0 × 10^{8})/_{( 8.0 × 108 )} = 1.50  The angle of refraction in glass for a ray of light incident at the airglass interface at an angle of incidence of 40^{o}.
Sin 40^{o}/_{sin r} =1.50
r = sin^{1}(^{sin40}/_{1.50})= 10.4^{o}.
 The refractive index of glass.
 Calculate the speed of light in diamond of refractive index 8.1.
η_{d} =velocity of light in vacuum/velocity of light in diamond
8.4= (3.0 × 10^{8})/V_{d}V_{d} =(3.0 × 10^{8} )/_{8.4} =1.25 × 2 8 m/s.  The speed of light in medium 1 is 8.0 × 10^{8} m/s and in medium 2 is 1.5 × 10^{8} m/s. Calculate the refractive index of medium 2 with respect to medium 1.
1η2 =^{V1}/V_{2} = (8.0 × 10^{8} m/s)/_{(1.5 × 108 m/s)}=1.33
Total Internal Reflection, Critical Angle and Refractive index
 As the angle of incidence in the denser medium increases the angle of refraction also increases. If this continues until the angle of refraction reaches 90^{o} , the angle of incidence is called the critical angle C. A critical angle is defined as the angle of incidence in the denser medium for which the angle of refraction is 90^{o} in the less dense medium.
 By the principle of the reversibility of light,
_{a}η_{g} = ^{sin90}/_{sin C} =^{1}/_{sin C.}  If the angle of incidence exceeds the critical angle, the light undergoes total internal reflection. This reflection obeys all the laws of reflection.
 For total internal reflection to occur, two conditions must be satisfied, namely:
â Light must pass from an optically denser medium to a less optically dense medium.
â The angle of incidence in the denser medium must be greater than the critical angle.
Example 8.5