In the previous section we saw dispersion of light. In this section, we will see formation of rainbow.
• For the formation of a rainbow in the sky, three conditions are to be satisfied. Let us see them first:
1.A rainbow is formed when the following two items are present at the same time in the atmosphere:
(i) sun light (ii) water particles
2. A rainbow is formed either in the morning or in the evening.
• Note that, at these times, the light from the sun are nearly parallel to the surface of the earth
3. When the sun is in the east, the rainbow is in the west
• When the sun is in the west, the rainbow is in the east
• So in the morning time, we must look towards the west to see the rainbow
• And in the evening time, we must look towards the east to see the rainbow
• In other words, to see the rainbow, our backs must be facing the sun
Above three are the conditions. Now we will see two properties of the rainbow
1. When viewed from the ground, a rainbow is always seen in the form of an arc. See image below:
• But when viewed from an aeroplane, a rainbow can be seen as a full circle. Some images can be seen here.
2. In a rainbow, red will be always at the outer edge
• Voilet will always be at the inner edge
Now we will discuss how a rainbow is formed. During this discussion, we will see the explanations for the above 3 conditions and the 2 properties. We will write the steps:
1. The water particles in the atmosphere will be spherical in shape.
• They can be considered as perfect spheres.
• Note that, they are not bubbles with air trapped inside. They are solid spheres, made up of water.
2. One such sphere is shown in fig.13.3(a) below:
• P is a point on the sphere. A white ray of light is incident at P
• A ray of light can fall on the surface of a sphere in two ways:
♦ It can fall obliquely on the surface
♦ It can fall normally on the surface
3. If a ray falls 'normally on the surface of a water sphere',
♦ it will pass (with out any deviation) through the centre of the sphere.
♦ also it will emerge without any deviation from the other side of the sphere.
■ When do we say that 'a ray falls normally on the surface of a sphere'?
Ans:
(i) Consider the green line drawn at 'P'
(ii) It is a radial line. That is., if the green line is extended downwards, it will pass through the centre of the sphere
(iii) We can easily draw this green line by joining 'P' and the 'centre'
(iv) This green line is the normal to the surface of the sphere at p.
(v) If a light ray falls along this normal, we say: 'that ray falls normally on the surface of a sphere'
4. Thus, in our present case, it is obvious that, the ray under consideration is falling obliquely at P.
• Because it is not falling along the normal at P
• Since it falls obliquely, refraction takes place. We have seen the result of such a refraction in fig.13.1 in the previous section
5. So the white light is split into it's component colours.
• We will take the two extreme colours: Red and violet. This is shown in fig.13.3(b)
♦ The red ray falls at Q on the inside surface of the sphere
♦ The violet ray falls at R on the inside surface of the sphere
6. Now consider fig.13.4(a) below:
• Green normals are drawn at Q and R
(i) Carefully consider the angle between the following two lines:
♦ The green normal at Q
♦ The line PQ
• The angle between the above two lines, is the 'angle of incidence at Q'
• This angle is greater than the critical angle
♦ So ray PQ cannot emerge out of the sphere. It will be reflected back
(We have learned about critical angle and total internal reflection here)
(ii) Again carefully consider the angle between the following two lines:
♦ The green normal at R
♦ The line PR
• The angle between the above two lines, is the 'angle of incidence at R'
• This angle is also greater than the critical angle
♦ So ray PR cannot emerge out of the sphere. It will also be reflected back
• The other colours which fall in between red and violet will also be reflected back
7. After reflection at Q, the red ray falls at S
• After reflection at R, the violet ray falls at T
• This time the angle of incidence at S is less than the critical angle
♦ So the red ray emerges out
• The angle of incidence at T is also less than the critical angle
♦ So the violet ray also emerges out
8. Thus, in effect, we have these:
• A white ray going into a water drop
• It comes out of the water drop as seven different colours.
• Red at one edge and violet at the other edge.
This is shown in fig.13.4(b)
• The other 5 colours fall in between red and violet in regular order.
9. Note that, in the case of a glass prism, we get the component colours on the other side of the prism
♦ With red at top and violet at bottom
• But in the case of a water drop, we get the component colours on the same side
♦ With red at bottom and violet at top
10. The resulting red ray will be always at an 'angle of deviation' of 42 from the original white ray
• The resulting violet ray will be always at an 'angle of deviation' of 40 from the original white ray
This is shown in fig.13.4(b)
■ 42 is greater than 40. So, when the refraction is done by a water drop in this way:
• Red will always be at bottom and
• Violet will always be at top
11. Now consider a person viewing the rainbow. This is shown in fig.13.5 below:
• Consider the line joining the eye of the viewer to the top most red arc of the rainbow
• One water drop along that line is shown in the fig.
• All seven colours will be emerging from that drop
• But red is the bottom most colour. Since all the other six colours are above red, they will travel above the head of the observer
• All water drops along that line will be producing all the seven colours
• But, since red is at the bottom, the colours other than red, from all of those drops, will pass above the head
• The observer will not see those colours
• That is the reason why red is always seen at the outer most arc in the rainbow
12. Now consider the line joining the eye of the viewer to the bottom most violet arc of the rainbow
• One water drop along that line is shown in the fig.
• All seven colours will be emerging from that drop
• But violet is the top most colour. Since all the other six colours are below violet, they will travel towards the feet of the observer
• All water drops along that line will be producing all the seven colours
• But, since violet is at the top, the colours other than violet, from all of those drops, will travel towards the feet
• The observer will not see those colours
• That is the reason why violet is always seen at the inner most arc in the rainbow
13. There are large numbers of water drops. All of them are dispersing light in different directions
• But those 'drops capable of sending particular colours to the viewer' lies in a circle around the viewer
• So we see a circular shape
14. The line from the eye to the red arc (in fig.13.5) will be a sloping line because red arc will be above the eye level
• The line from the eye to the violet arc (in fig.13.5) will also be a sloping line because violet arc will also be above the eye level
• But the line from the eye to the point 'O' will be a horizontal line.
15. Part of the circle below the horizon will not be visible. Also, there are no water particles below the horizon to disperse the light
• So we see only part of the circle. That is., an arc.
• But when viewed from an aeroplane, water particles are available below the horizontal line. So the full circle will be visible
• For the formation of a rainbow in the sky, three conditions are to be satisfied. Let us see them first:
1.A rainbow is formed when the following two items are present at the same time in the atmosphere:
(i) sun light (ii) water particles
2. A rainbow is formed either in the morning or in the evening.
• Note that, at these times, the light from the sun are nearly parallel to the surface of the earth
3. When the sun is in the east, the rainbow is in the west
• When the sun is in the west, the rainbow is in the east
• So in the morning time, we must look towards the west to see the rainbow
• And in the evening time, we must look towards the east to see the rainbow
• In other words, to see the rainbow, our backs must be facing the sun
Above three are the conditions. Now we will see two properties of the rainbow
1. When viewed from the ground, a rainbow is always seen in the form of an arc. See image below:
 |
| Source: By Eric Rolph at English Wikipedia - English Wikipedia, CC BY-SA 2.5, https://commons.wikimedia.org/w/index.php?curid=2406447 |
2. In a rainbow, red will be always at the outer edge
• Voilet will always be at the inner edge
Now we will discuss how a rainbow is formed. During this discussion, we will see the explanations for the above 3 conditions and the 2 properties. We will write the steps:
1. The water particles in the atmosphere will be spherical in shape.
• They can be considered as perfect spheres.
• Note that, they are not bubbles with air trapped inside. They are solid spheres, made up of water.
2. One such sphere is shown in fig.13.3(a) below:
 |
| Fig.13.3 |
• A ray of light can fall on the surface of a sphere in two ways:
♦ It can fall obliquely on the surface
♦ It can fall normally on the surface
3. If a ray falls 'normally on the surface of a water sphere',
♦ it will pass (with out any deviation) through the centre of the sphere.
♦ also it will emerge without any deviation from the other side of the sphere.
■ When do we say that 'a ray falls normally on the surface of a sphere'?
Ans:
(i) Consider the green line drawn at 'P'
(ii) It is a radial line. That is., if the green line is extended downwards, it will pass through the centre of the sphere
(iii) We can easily draw this green line by joining 'P' and the 'centre'
(iv) This green line is the normal to the surface of the sphere at p.
(v) If a light ray falls along this normal, we say: 'that ray falls normally on the surface of a sphere'
4. Thus, in our present case, it is obvious that, the ray under consideration is falling obliquely at P.
• Because it is not falling along the normal at P
• Since it falls obliquely, refraction takes place. We have seen the result of such a refraction in fig.13.1 in the previous section
5. So the white light is split into it's component colours.
• We will take the two extreme colours: Red and violet. This is shown in fig.13.3(b)
♦ The red ray falls at Q on the inside surface of the sphere
♦ The violet ray falls at R on the inside surface of the sphere
6. Now consider fig.13.4(a) below:
 |
| Fig.13.4 |
(i) Carefully consider the angle between the following two lines:
♦ The green normal at Q
♦ The line PQ
• The angle between the above two lines, is the 'angle of incidence at Q'
• This angle is greater than the critical angle
♦ So ray PQ cannot emerge out of the sphere. It will be reflected back
(We have learned about critical angle and total internal reflection here)
(ii) Again carefully consider the angle between the following two lines:
♦ The green normal at R
♦ The line PR
• The angle between the above two lines, is the 'angle of incidence at R'
• This angle is also greater than the critical angle
♦ So ray PR cannot emerge out of the sphere. It will also be reflected back
• The other colours which fall in between red and violet will also be reflected back
7. After reflection at Q, the red ray falls at S
• After reflection at R, the violet ray falls at T
• This time the angle of incidence at S is less than the critical angle
♦ So the red ray emerges out
• The angle of incidence at T is also less than the critical angle
♦ So the violet ray also emerges out
8. Thus, in effect, we have these:
• A white ray going into a water drop
• It comes out of the water drop as seven different colours.
• Red at one edge and violet at the other edge.
This is shown in fig.13.4(b)
• The other 5 colours fall in between red and violet in regular order.
9. Note that, in the case of a glass prism, we get the component colours on the other side of the prism
♦ With red at top and violet at bottom
• But in the case of a water drop, we get the component colours on the same side
♦ With red at bottom and violet at top
10. The resulting red ray will be always at an 'angle of deviation' of 42 from the original white ray
• The resulting violet ray will be always at an 'angle of deviation' of 40 from the original white ray
This is shown in fig.13.4(b)
■ 42 is greater than 40. So, when the refraction is done by a water drop in this way:
• Red will always be at bottom and
• Violet will always be at top
11. Now consider a person viewing the rainbow. This is shown in fig.13.5 below:
 |
| Fig.13.5 |
• One water drop along that line is shown in the fig.
• All seven colours will be emerging from that drop
• But red is the bottom most colour. Since all the other six colours are above red, they will travel above the head of the observer
• All water drops along that line will be producing all the seven colours
• But, since red is at the bottom, the colours other than red, from all of those drops, will pass above the head
• The observer will not see those colours
• That is the reason why red is always seen at the outer most arc in the rainbow
12. Now consider the line joining the eye of the viewer to the bottom most violet arc of the rainbow
• One water drop along that line is shown in the fig.
• All seven colours will be emerging from that drop
• But violet is the top most colour. Since all the other six colours are below violet, they will travel towards the feet of the observer
• All water drops along that line will be producing all the seven colours
• But, since violet is at the top, the colours other than violet, from all of those drops, will travel towards the feet
• The observer will not see those colours
• That is the reason why violet is always seen at the inner most arc in the rainbow
13. There are large numbers of water drops. All of them are dispersing light in different directions
• But those 'drops capable of sending particular colours to the viewer' lies in a circle around the viewer
• So we see a circular shape
14. The line from the eye to the red arc (in fig.13.5) will be a sloping line because red arc will be above the eye level
• The line from the eye to the violet arc (in fig.13.5) will also be a sloping line because violet arc will also be above the eye level
• But the line from the eye to the point 'O' will be a horizontal line.
15. Part of the circle below the horizon will not be visible. Also, there are no water particles below the horizon to disperse the light
• So we see only part of the circle. That is., an arc.
• But when viewed from an aeroplane, water particles are available below the horizontal line. So the full circle will be visible
In the next section, we will see recombination of light.
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