In the previous section we saw the basics about refraction. In this section we will see index of refraction.
• Two experiments were conducted. One with glass and other with water
• Fig.7.5 below shows the results of two experiments
• The details are given below:
Fig.7.5(a):
1. Light was passed from air to glass
2. The angle of incidence was 30o.
3. The angle of refraction in glass was noted down. Let it be rg
Fig.7.5(b):
1. Light was passed from air to water
2. The angle of incidence was again 30o.
3. The angle of refraction in water was noted down. Let it be rw
■ rg and rw were compared with each other. It was found that rg < rw.
• That is., the ray of light inside the glass deviates more towards the normal
• Note that, in both the cases, the refracted rays deviates towards the normal
♦ This is because, in both the cases, the light pass from rarer to denser
■ But in glass, the refracted ray is deviated more. What could be the reason?
• To give an explanation, we must first learn about a special ratio.
• This ratio is called Refractive index.
■ Refractive index of a medium is the ratio of the following two quantities:
(i) Speed of light in vacuum
(ii) Speed of light in that medium
■ So we can write:
n = c⁄v
Where,
n = refractive index of the medium under consideration
c = speed of light in vacuum
v = speed of light in that medium under consideration
■ So if we want to find the 'n' value of a material,
• we must first find 'v', which is the speed of light in that medium
• Then divide c by that v. We will get the 'n' of that medium.
■ The speed of light in vacuum is almost equal to the 'speed of light in air'.
• So instead of 'c', we can use 'speed of light in air'
• 'Speed of light in air' is already calculated by scientists. It is 3.0 × 108 m/s
■ So we can write:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
• Scientists have already calculated the 'n' of many common materials. A list can be seen here in wikipedia.
An example:
Speed of light in glass is 2.0 × 108 m/s. Calculate the refractive index 'n' of glass
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. Substituting the values, we get:
n = (3.0 × 108)⁄(2.0 × 108) = 3⁄2 = 1.5
• Note that, 'n' is a ratio. So it has no units
Another example:
Speed of light in water is 2.25 × 108 m/s. Calculate the refractive index 'n' of water
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. Substituting the values, we get:
n = (3.0 × 108)⁄(2.25 × 108) = 3⁄2.25 = 1.33
One more example:
In an aquarium, light at the speed of 3.0 × 108 m/s is incident obliquely on the surface of water. If the refractive index 'n' of water is 1.33, find out the speed of light in water.
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. Substituting the known values, we get:
1.33 = (3.0 × 108)⁄v ⟹ v = (3.0 × 108)⁄1.33 = 2.25 × 108 m/s
Solved example 7.1
Refractive indices of some materials are given below. Find out through which medium light passes with maximum speed
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. The numerator is 'Speed of light in air'. It will be the same for all the cases
• 'v' is in the denominator.
• In any ratio, if the denominator is to be maximum, the quotient 'n' has to be minimum
3. The smallest quotient in the given table is 1.33
• So we can write:
Among the given materials, 'speed of light' will be maximum in water.
Solved example 7.2
Observe the fig.7.6 below. Light falling on two different media are shown.
(a) Which medium has greater optical density
(b) Why?
(c) Which medium has greater refractive index?
Solution:
Part (a):
Medium 1 has greater optical density
Part (b):
1. In both cases the refracted rays deviates towards the normal
2. This deviation is more for medium 1
3. That is because, light finds it more difficult to travel in medium 1 than in medium 2
4. So medium 1 has greater optical density
Part (c):
1. Medium 1 has greater optical density. So velocity in medium 1 will be lesser than in medium 2
That is: v1 < v2
2. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
3. The numerator is 'Speed of light in air'. It will be the same for both the cases
• 'v' is in the denominator.
• In any ratio, if the denominator is smaller, the quotient 'n' will be higher
• So the quotient will be higher for medium 1
Thus we can write:
■ Medium 1 has greater refractive index
1. Consider fig.7.7(a) below. A ray of light is passing from water to air
• Water is denser than air. So the light is passing from a denser medium to a rarer medium
♦ Then the refracted ray will deviate away from the normal
♦ We can see that this is indeed true in fig.7.7(a)
• The angle of incidence is 30o. The angle of refraction is marked as r1.
2. Now the angle of incidence is gradually increased. It is made 40o. This is shown in fig.7.7(b)
• The angle of refraction also increases. It is marked as r2.
3. So, if we increase the angle of incidence, the angle of refraction will also increase
• This means that the refracted ray gets closer and closer to the boundary between air and water
4. If we go on increasing the angle of incidence, a stage will be reached in which the refracted ray passes exactly along the boundary.
• At this stage the angle of refraction is 90o.
5. When light passes from water to air, this stage occurs when the angle of incidence is 48.6o.
• This angle is known as the critical angle of water. It is shown in fig.7.7(c)
6. If the angle of incidence is greater than the critical angle, the light will not pass into the air. It will be reflected back into the water.
• This reflection will occur at the boundary
• That is., the boundary between air and water will act as a mirror.
7. In fig.d, the angle of incidence is 55o. The light will not pass into the air. It will be reflected back
Another experiment:
1. Consider fig.7.8(a) below. A ray of light is passing from glass to air
• Glass is denser than air. So the light is passing from a denser medium to a rarer medium
♦ Then the refracted ray will deviate away from the normal
♦ We can see that this is indeed true in fig.7.8(a)
• The angle of incidence is 30o. The angle of refraction is marked as r1.
2. Now the angle of incidence is gradually increased. It is made 40o. This is shown in fig.7.8(b)
• The angle of refraction also increases. It is marked as r2.
3. So, if we increase the angle of incidence, the angle of refraction will also increase
• This means that the refracted ray gets closer and closer to the boundary between air and glass
4. If we go on increasing the angle of incidence, a stage will be reached in which the refracted ray passes exactly along the boundary.
• At this stage the angle of refraction is 90o.
5. When light passes from glass to air, this stage occurs when the angle of incidence is 42o.
• This angle is known as the critical angle of glass. It is shown in fig.7.8(c)
6. If the angle of incidence is greater than the critical angle, the light will not pass into the air. It will be reflected back into the glass.
• This reflection will occur at the boundary
• That is., the boundary between air and glass will act as a mirror.
7. In fig.d, the angle of incidence is 45o. The light will not pass into the air. It will be reflected back
Critical angle:
When a ray of light passes from a medium of greater optical density to that of a lower optical density, the angle of incidence at which the angle of refraction becomes 90o is the critical angle.
Total internal reflection:
When a ray of light passes from a medium of greater optical density to that of a lower optical density, at an angle of incidence greater than the critical angle, the ray is reflected back to the same medium without undergoing refraction. This phenomenon is called total internal reflection.
Practical applications of total internal reflection:
1. Optical fibre
Optical fibres are thin bendable fibres. Their structure is shown in fig.7.9 below.
• They are made of glass or transparent plastic.
• Their size is similar to the size of a hair. See images here.
1. Consider the first incident ray entering the tube. It travels through the blue layer and tries to get out of the tube.
2. But it falls on the white layer. This white layer is rarer than the blue.
3. So the light gets reflected back completely into the blue layer.
4. This reflected light hits the other side of the tube. There also it is reflected back.
5. Thus the light is continuously reflected back and thus reaches the other end of the tube
Comparison between optical fibres and metallic conductors:
1. Data in the form of electrical signals can be converted into light signals. They can then be sent to any distance with out energy and intensity loss
2. Optical fibres are not affected by lightning. But it is better to consult the manufactures and take precautionary measures.
3. Optical fibres are not affected by acid or alkali. So they do not get damaged in the soil
4. It needs less maintenance. So it is cost effective
5. Large number of signals of different frequencies can be sent at a time
2. Endoscopy
An endoscope consists of a flexible tube with a light and camera attached to it. Using it, doctors can view the pictures of the digestive tract on a color monitor.
3. Electric decoration
In many fancy lights, optical fibres are used to make different colored patterns.
• Two experiments were conducted. One with glass and other with water
• Fig.7.5 below shows the results of two experiments
 |
| Fig.7.5 |
Fig.7.5(a):
1. Light was passed from air to glass
2. The angle of incidence was 30o.
3. The angle of refraction in glass was noted down. Let it be rg
Fig.7.5(b):
1. Light was passed from air to water
2. The angle of incidence was again 30o.
3. The angle of refraction in water was noted down. Let it be rw
■ rg and rw were compared with each other. It was found that rg < rw.
• That is., the ray of light inside the glass deviates more towards the normal
• Note that, in both the cases, the refracted rays deviates towards the normal
♦ This is because, in both the cases, the light pass from rarer to denser
■ But in glass, the refracted ray is deviated more. What could be the reason?
• To give an explanation, we must first learn about a special ratio.
• This ratio is called Refractive index.
■ Refractive index of a medium is the ratio of the following two quantities:
(i) Speed of light in vacuum
(ii) Speed of light in that medium
■ So we can write:
n = c⁄v
Where,
n = refractive index of the medium under consideration
c = speed of light in vacuum
v = speed of light in that medium under consideration
■ So if we want to find the 'n' value of a material,
• we must first find 'v', which is the speed of light in that medium
• Then divide c by that v. We will get the 'n' of that medium.
■ The speed of light in vacuum is almost equal to the 'speed of light in air'.
• So instead of 'c', we can use 'speed of light in air'
• 'Speed of light in air' is already calculated by scientists. It is 3.0 × 108 m/s
■ So we can write:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
• Scientists have already calculated the 'n' of many common materials. A list can be seen here in wikipedia.
An example:
Speed of light in glass is 2.0 × 108 m/s. Calculate the refractive index 'n' of glass
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. Substituting the values, we get:
n = (3.0 × 108)⁄(2.0 × 108) = 3⁄2 = 1.5
• Note that, 'n' is a ratio. So it has no units
Another example:
Speed of light in water is 2.25 × 108 m/s. Calculate the refractive index 'n' of water
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. Substituting the values, we get:
n = (3.0 × 108)⁄(2.25 × 108) = 3⁄2.25 = 1.33
One more example:
In an aquarium, light at the speed of 3.0 × 108 m/s is incident obliquely on the surface of water. If the refractive index 'n' of water is 1.33, find out the speed of light in water.
Solution:
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. Substituting the known values, we get:
1.33 = (3.0 × 108)⁄v ⟹ v = (3.0 × 108)⁄1.33 = 2.25 × 108 m/s
Solved example 7.1
Refractive indices of some materials are given below. Find out through which medium light passes with maximum speed
| Medium | Refractive index |
|---|---|
| Glass | 1.52 |
| Glycerine | 1.47 |
| Sunflower oil | 1.47 |
| Water | 1.33 |
| Flint glass | 1.62 |
1. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
2. The numerator is 'Speed of light in air'. It will be the same for all the cases
• 'v' is in the denominator.
• In any ratio, if the denominator is to be maximum, the quotient 'n' has to be minimum
3. The smallest quotient in the given table is 1.33
• So we can write:
Among the given materials, 'speed of light' will be maximum in water.
Solved example 7.2
Observe the fig.7.6 below. Light falling on two different media are shown.
 |
| Fig.7.6 |
(b) Why?
(c) Which medium has greater refractive index?
Solution:
Part (a):
Medium 1 has greater optical density
Part (b):
1. In both cases the refracted rays deviates towards the normal
2. This deviation is more for medium 1
3. That is because, light finds it more difficult to travel in medium 1 than in medium 2
4. So medium 1 has greater optical density
Part (c):
1. Medium 1 has greater optical density. So velocity in medium 1 will be lesser than in medium 2
That is: v1 < v2
2. We have:
n = Speed of light in air⁄v
Where,
n = refractive index of the medium under consideration
v = speed of light in that medium under consideration
3. The numerator is 'Speed of light in air'. It will be the same for both the cases
• 'v' is in the denominator.
• In any ratio, if the denominator is smaller, the quotient 'n' will be higher
• So the quotient will be higher for medium 1
Thus we can write:
■ Medium 1 has greater refractive index
Total internal reflection
Let us see an experiment. We will write the details in steps:1. Consider fig.7.7(a) below. A ray of light is passing from water to air
 |
| Fig.7.7 |
♦ Then the refracted ray will deviate away from the normal
♦ We can see that this is indeed true in fig.7.7(a)
• The angle of incidence is 30o. The angle of refraction is marked as r1.
2. Now the angle of incidence is gradually increased. It is made 40o. This is shown in fig.7.7(b)
• The angle of refraction also increases. It is marked as r2.
3. So, if we increase the angle of incidence, the angle of refraction will also increase
• This means that the refracted ray gets closer and closer to the boundary between air and water
4. If we go on increasing the angle of incidence, a stage will be reached in which the refracted ray passes exactly along the boundary.
• At this stage the angle of refraction is 90o.
5. When light passes from water to air, this stage occurs when the angle of incidence is 48.6o.
• This angle is known as the critical angle of water. It is shown in fig.7.7(c)
6. If the angle of incidence is greater than the critical angle, the light will not pass into the air. It will be reflected back into the water.
• This reflection will occur at the boundary
• That is., the boundary between air and water will act as a mirror.
7. In fig.d, the angle of incidence is 55o. The light will not pass into the air. It will be reflected back
Another experiment:
1. Consider fig.7.8(a) below. A ray of light is passing from glass to air
 |
| Fig.7.8 |
♦ Then the refracted ray will deviate away from the normal
♦ We can see that this is indeed true in fig.7.8(a)
• The angle of incidence is 30o. The angle of refraction is marked as r1.
2. Now the angle of incidence is gradually increased. It is made 40o. This is shown in fig.7.8(b)
• The angle of refraction also increases. It is marked as r2.
3. So, if we increase the angle of incidence, the angle of refraction will also increase
• This means that the refracted ray gets closer and closer to the boundary between air and glass
4. If we go on increasing the angle of incidence, a stage will be reached in which the refracted ray passes exactly along the boundary.
• At this stage the angle of refraction is 90o.
5. When light passes from glass to air, this stage occurs when the angle of incidence is 42o.
• This angle is known as the critical angle of glass. It is shown in fig.7.8(c)
6. If the angle of incidence is greater than the critical angle, the light will not pass into the air. It will be reflected back into the glass.
• This reflection will occur at the boundary
• That is., the boundary between air and glass will act as a mirror.
7. In fig.d, the angle of incidence is 45o. The light will not pass into the air. It will be reflected back
Critical angle:
When a ray of light passes from a medium of greater optical density to that of a lower optical density, the angle of incidence at which the angle of refraction becomes 90o is the critical angle.
Total internal reflection:
When a ray of light passes from a medium of greater optical density to that of a lower optical density, at an angle of incidence greater than the critical angle, the ray is reflected back to the same medium without undergoing refraction. This phenomenon is called total internal reflection.
Practical applications of total internal reflection:
1. Optical fibre
Optical fibres are thin bendable fibres. Their structure is shown in fig.7.9 below.
• They are made of glass or transparent plastic.
• Their size is similar to the size of a hair. See images here.
 |
| Fig.7.9 |
2. But it falls on the white layer. This white layer is rarer than the blue.
3. So the light gets reflected back completely into the blue layer.
4. This reflected light hits the other side of the tube. There also it is reflected back.
5. Thus the light is continuously reflected back and thus reaches the other end of the tube
Comparison between optical fibres and metallic conductors:
1. Data in the form of electrical signals can be converted into light signals. They can then be sent to any distance with out energy and intensity loss
2. Optical fibres are not affected by lightning. But it is better to consult the manufactures and take precautionary measures.
3. Optical fibres are not affected by acid or alkali. So they do not get damaged in the soil
4. It needs less maintenance. So it is cost effective
5. Large number of signals of different frequencies can be sent at a time
2. Endoscopy
An endoscope consists of a flexible tube with a light and camera attached to it. Using it, doctors can view the pictures of the digestive tract on a color monitor.
3. Electric decoration
In many fancy lights, optical fibres are used to make different colored patterns.
