In the previous section we saw lens formula and magnification. In this section, we will learn about the Eye and vision.
The structure of the eye is shown in the fig. below. It is obtained from wikimedia commons.
Let us see the important features:
1. The amount of light reaching the lens is controlled by the iris.
• When the surrounding light is less, the iris expands to let in more light which may be required for clear vision
• When the surrounding light is more, the iris contracts to let in only the just required light for clear vision
2. The lens in the front of the eye is a convex lens
• So light rays emerging from an object will have to pass through this convex lens
• When they pass through the lens, they will converge. This is shown schematically in the fig.7.28 below:
[Note that, in the fig.7.28, two light rays emerging from a 'point object' is shown. This is for convenience only. For larger objects, more rays will have to be drawn]
3. In fig.7.28, we can see that the image is formed exactly on the rear side of the eyeball.
• At this rear side, the retina is present. This is shown in yellow colour in the previous fig.7.27 above.
• So we can say, the image is formed exactly on the retina.
• Optical nerves of the retina send the sense of vision to the brain. Due to the activity of brain, we see the object
4. We now know how we see 'O'.
■ But what if the position of 'O' change?
• The lens cannot change it's position
• The retina cannot change it's position
• So the only option is to 'change the focal length' of the lens
♦ 'Change in focal length' is achieved by 'changing the curvature' of the lens
♦ 'Change in curvature' is achieved by the expansion and contraction of the ciliary muscles
• Thus we can say:
The eye has the ability to form an image, whatever be the position of the object
The ability of the eye to form an image on the retina by adjusting the focal length of the lens in the eye, by varying the curvature of the lens whatever be the position of the object, is the power of accommodation
• If we hold a book very close to our eyes, we will not be able to read the letters.
• That means, in spite of the 'power of accommodation', there are some limitations.
• If the letters are too close, a clear image (as shown in fig.7.28) cannot be formed.
• The lens cannot accommodate such an extreme condition
■ The nearest point at which one can see an object clearly is called the Near point. The near point of a healthy eye is 25 cm
• In the same way, in the case of 'very far away objects' also, there are limitations
Consider fig.7.29 below. We will do an analysis of the fig.:
1. In fig.7.29(a), after passing through the lens, the rays are indeed converging.
• But the point of convergence is not on the retina
• The point of convergence is behind the retina. What could be the reason for this?
2. It may be due to any one or both of the two reasons given below:
(i) The power of the lens decreased
• 'Decrease in the power' of lens means that the lens is not able to 'cause enough bending' of light rays
• In the fig.7.29(a) we see that, if the lens is able to cause a little more bending, the light rays would surely intersect at the retina
(ii) The eyeball contracted so that the retina moved forward
3. What will be the result when such defects occur?
• We see that a perfect image will be formed behind the retina.
• But that image has no use. The retina can send signals based on only those images which fall exactly on it
■ We see that rays 'nearly' converge on the retina. But the convergence is not perfect.
• So the image formed on the retina will not be perfect. It will be 'blurred'.
• So a person with such defects will see only a blurred image
4. What is the remedy for this defect?
Consider fig.7.29(b). After passing through the lens, two sets of rays are shown
(i) A set consisting of two white rays
♦ This is the same shown in fig.a
(ii) A set consisting of two green rays
• If we follow the white set, we can see that the problem will not be remedied.
♦ They intersect behind the retina
• If we follow the green set, we can see that the problem will be remedied.
♦ They intersect exactly on the retina
5. So we want the green set. That is., after passing through the lens, the rays must follow the green lines. How can we make that happen?
• We notice that, the green rays 'converge more' than the white rays. That is why they are able to intersect exactly on the retina
• So if we can make the rays to 'converge more', the defect will be remedied
6. How can we make them to 'converge more'?
Ans: By using a converging lens.
A convex lens is a converging lens. So we can use it. It is shown in fig.7.29(c)
7. So we must put a convex lens in front of the eye to remedy this defect
• The convex lens in front of the eye, will create some degree of convergence even before the light enters the eye
• Then the lens in the eye will need to do only a 'lesser work' for making the rays intersect exactly on the retina
• The power of the 'convex lens to be used' will be prescribed by the doctor after examination
Case 2:
Consider fig.7.30 below. We will do an analysis of the fig.:
1. In fig.7.30(a), after passing through the lens, the rays are indeed converging.
• But the point of convergence is not on the retina
• The point of convergence is in front of the retina. What could be the reason for this?
2. It may be due to any one or both of the two reasons given below:
(i) The lens has increased power
• 'Increase in the power' of lens means that the lens 'cause a bending, which is more than what is required'.
(ii) The eyeball is longer than normal so that the retina moved backwards
3. What will be the result when such defects occur?
• We see that a perfect image will be formed in front of the retina.
• But that image has no use. The retina can send signals based on only those images which fall exactly on it
■ We see that rays 'nearly' converge on the retina. But the convergence is not perfect.
• So the image formed on the retina will not be perfect. It will be 'blurred'.
• So a person with such defects will see only a blurred image
4. What is the remedy for this defect?
Consider fig.7.30(b). After passing through the lens, two sets of rays are shown
(i) A set consisting of two white rays
♦ This is the same shown in fig.a
(ii) A set consisting of two green rays
• If we follow the white set, we can see that the problem will not be remedied.
♦ They intersect in front of the retina
• If we follow the green set, we can see that the problem will be remedied.
♦ They intersect exactly on the retina
5. So we want the green set. That is., after passing through the lens, the rays must follow the green lines. How can we make that happen?
• We notice that, the green rays 'converge less' than the white rays. That is why they are able to intersect exactly on the retina
• So if we can make the rays to 'converge less', the defect will be remedied
6. How can we make them to 'converge less'?
Ans: 'Converging less' is equivalent to 'diverging more'
• So, by using a diverging lens, we can remedy this defect.
• A concave lens is a diverging lens. So we can use it
7. So we must put a concave lens in front of the eye to remedy this defect
• The concave lens in front of the eye, will create some degree of divergence even before the light enters the eye
• Then the rays will intersect exactly on the retina, even if the lens of the eye has increased power
• The power of the 'concave lens to be used' will be prescribed by the doctor after examination
The two defects that we saw above are given special names:
Case 1:
This is called Hypermetropia or Far-sightedness
The definition is:
■ Since the image is formed behind the retina, instead of being formed at the retina, even though distant objects are clearly seen, nearer objects cannot be seen. This defect of the eye is called Hypermetropia or Far-sightedness
• In the above definition, we see a peculiarity:
• It is said that, 'distant objects can be clearly seen'. That means people with hypermetropia:
♦ Can see distant objects
♦ Cannot see near objects
• How is that possible?
We will write the answer in steps:
1. Consider the fig.7.31(a) below:
• It shows an eye with out any defects.
• The white rays from the near object O are neatly converging on the retina
2. Another set of green rays are shown in the same fig.a
• Those rays are shown to be parallel to each other.
3. Why are they parallel to each other?
• The answer is that, those rays come from distant objects, and such rays will be always parallel to each other
4. In fig.7.31(b), two additional arcs are shown
• One magenta arc and one yellow arc
• The magenta arc shows the angle between incident white ray and refracted white ray
• The yellow arc shows the angle between incident green ray and refracted white ray
• Obviously, the angle indicated by the magenta arc is larger than that indicated by the yellow arc
5. So now compare the incident green set and the incident white set in fig.b
• We can see that, 'bending through a greater angle' is required to bring the white rays together on to the retina
• In other words, when compared to the white rays, a 'lesser effort' by the lens is sufficient to bring the green rays together on to the retina
6. Since 'lesser effort' is sufficient, even eyes with hypermetropia can make the rays from distant objects to intersect on the retina
• In other words, persons with hypermetropia will be able to see distant objects with out external help.
• While they cannot see near objects clearly. So another name for this defect is Far-sightedness.
Case 2:
This is called Myopia or Near-sightedness
The definition is:
■ Since the image is formed in front of the retina, instead of being formed at the retina, even though nearby objects are clearly seen, distant objects cannot be seen. This defect of the eye is called Myopia or Near-sightedness
• In the above definition, we see a peculiarity:
It is said that, 'near objects can be clearly seen'. That means people with myopia:
♦ Can see near objects
♦ Cannot see distant objects
• How is that possible?
The answer is just opposite to the one that we saw for case 1. But we will write the steps again:
1. Consider the same fig.7.31 above.
• It shows an eye with out any defects.
• The white rays from the near object O are neatly converging on the retina
2. Another set of green rays are shown in the same fig.a
• Those rays are shown to be parallel to each other.
3. Why are they parallel to each other?
• The answer is that, those rays come from distant objects, and such rays will be always parallel to each other
4. In fig.7.31(b), two additional arcs are shown
• One magenta arc and one yellow arc
• The magenta arc shows the angle between incident white ray and refracted white ray
• The yellow arc shows the angle between incident green ray and refracted white ray
• Obviously, the angle indicated by the magenta arc is larger than that indicated by the yellow arc
5. So now compare the green set and the white set.
• We can see that, 'more bending' is required to bring the white rays together on to the retina
• In other words, when compared to the white rays, a 'lesser effort' by the lens is sufficient to bring the green rays together on to the retina
6. Since 'lesser effort' is sufficient, the eyes with myopia (which has increased converging power) will be converging the green rays before they reach the retina.
• Converging of rays before reaching the retina is not useful
7. The white rays require 'more bending'. So the extra converging power available will be compensated.
• That is., the white rays will converge on the retina. Which means that near objects can be clearly seen without external help. So another name for this defect is Near-sightedness.
2. But if the curvature is not smooth, the surface of the lens will be irregular.
3. In such a situation, the light rays from both near and far objects will not be refracted properly.
4. They will not intersect on the retina. So both near and far objects will appear to be blurred.
5. This condition is called astigmatism.
6. This can be remedied by using cylindrical lens of appropriate power.
2. For this, the power of the lens of the eye will have to be increased.
• This is achieved with the help of the ciliary muscles
3. But in elderly people, the ciliary muscles may not be able to provide the necessary curvature to the lens.
4. So elderly people will have to hold the objects at a distance greater than 25 cm, to see them properly. This condition is called presbyopia.
5. This can be remedied by using convex lens of appropriate power
Doctor's prescription
• The prescription made by the doctor will show the power of the required lens for rectifying the defects of the eye.
• Power of a lens is related to it's focal length
• Let us see an example:
1. Focal length of a lens is 20 cm
2. Express this focal length in metres:
• We have: 20 cm = 0.20 m
3. Take the reciprocal
• We have: Reciprocal of 0.20 = 1⁄0.02 = 100⁄20 = 5
4. This '5' is the power of the lens.
We can write the definition of power:
■ Power of a lens is the reciprocal of focal length expressed in metres.
• Power = 1⁄f
♦ Where f is the focal length expressed in metres
• Unit of power is dioptre. It is represented by D
■ Power of convex lens is positive and that of a concave lens is negative
But why do we take the reciprocal of the focal length?
The answer can be written in steps:
1. Consider fig.7.13 that we saw in a previous section. For convenience, it is shown again below:
2. Consider the convex lens in fig.a. It is clear that convex lens is used for converging light rays
• If the convex lens has greater ability to converge rays, we can say that it has greater power
• If the convex lens has lesser ability to converge rays, we can say that it has lesser power
3. But from the fig.a, it is clear that:
• If greater convergence is achieved, the focus F will move towards the lens
♦ That is., focal length will decrease
• If lesser convergence is achieved, the focus F will move away from the lens
♦ That is., focal length will increase
4. So we can conclude. In the case of a convex lens:
• Greater power means lesser focal length
• Lesser power means greater focal length
5. So there is an 'inverse relation'
• Such a relation can be expressed only if we put 'f' in the denominator.
• Because in the relation Power = 1⁄f,
♦ When f increases, power decreases
♦ When f decreases, power increases
■ So we indeed have to take the reciprocal of 'f'
6. Now consider the concave lens in fig.b. It is clear that concave lens is used for diverging light rays
• If the concave lens has greater ability to diverge rays, we can say that it has greater power
• If the concave lens has lesser ability to diverge rays, we can say that it has lesser power
7. But from the fig.b, it is clear that:
• If greater divergence is achieved, the focus F will move towards the lens
♦ That is., focal length will decrease
• If lesser divergence is achieved, the focus F will move away from the lens
♦ That is., focal length will increase
8. So we can conclude. In the case of a concave lens:
• Greater power means lesser focal length
• Lesser power means greater focal length
9. So there is an 'inverse relation'
• Such a relation can be expressed only if we put 'f' in the denominator.
• Because in the relation Power = 1⁄f,
♦ When f increases, power decreases
♦ When f decreases, power increases
■ So for the concave lens also, we indeed have to take the reciprocal of 'f'
Solved example 7.6
The powers of the lenses in the prescription of the doctor are +1.50 D and +1 D. What are the focal lengths of these two lenses? What type of lenses are they?
Solution:
Case 1:
1. We have: Power = +1.5 D = 1⁄f
♦ Where f is the focal length expressed in metres
2. So f = 1⁄+1.5 = +10⁄15 = +2⁄3 = +0.666 metres = +66.6 cm
3. Since the focal length is positive, it is a convex lens
Case 2:
1. We have: Power = +1 D = 1⁄f
♦ Where f is the focal length expressed in metres
2. So f = 1⁄+1 = +1 metre = +100 cm
3. Since the focal length is positive, it is a convex lens
The structure of the eye is shown in the fig. below. It is obtained from wikimedia commons.
Let us see the important features:
1. The amount of light reaching the lens is controlled by the iris.
• When the surrounding light is less, the iris expands to let in more light which may be required for clear vision
• When the surrounding light is more, the iris contracts to let in only the just required light for clear vision
2. The lens in the front of the eye is a convex lens
• So light rays emerging from an object will have to pass through this convex lens
• When they pass through the lens, they will converge. This is shown schematically in the fig.7.28 below:
 |
| Fig.7.28 |
3. In fig.7.28, we can see that the image is formed exactly on the rear side of the eyeball.
• At this rear side, the retina is present. This is shown in yellow colour in the previous fig.7.27 above.
• So we can say, the image is formed exactly on the retina.
• Optical nerves of the retina send the sense of vision to the brain. Due to the activity of brain, we see the object
4. We now know how we see 'O'.
■ But what if the position of 'O' change?
• The lens cannot change it's position
• The retina cannot change it's position
• So the only option is to 'change the focal length' of the lens
♦ 'Change in focal length' is achieved by 'changing the curvature' of the lens
♦ 'Change in curvature' is achieved by the expansion and contraction of the ciliary muscles
• Thus we can say:
The eye has the ability to form an image, whatever be the position of the object
The ability of the eye to form an image on the retina by adjusting the focal length of the lens in the eye, by varying the curvature of the lens whatever be the position of the object, is the power of accommodation
• If we hold a book very close to our eyes, we will not be able to read the letters.
• That means, in spite of the 'power of accommodation', there are some limitations.
• If the letters are too close, a clear image (as shown in fig.7.28) cannot be formed.
• The lens cannot accommodate such an extreme condition
■ The nearest point at which one can see an object clearly is called the Near point. The near point of a healthy eye is 25 cm
• In the same way, in the case of 'very far away objects' also, there are limitations
Defects in vision
Case 1:Consider fig.7.29 below. We will do an analysis of the fig.:
 |
| Fig.7.29 |
• But the point of convergence is not on the retina
• The point of convergence is behind the retina. What could be the reason for this?
2. It may be due to any one or both of the two reasons given below:
(i) The power of the lens decreased
• 'Decrease in the power' of lens means that the lens is not able to 'cause enough bending' of light rays
• In the fig.7.29(a) we see that, if the lens is able to cause a little more bending, the light rays would surely intersect at the retina
(ii) The eyeball contracted so that the retina moved forward
3. What will be the result when such defects occur?
• We see that a perfect image will be formed behind the retina.
• But that image has no use. The retina can send signals based on only those images which fall exactly on it
■ We see that rays 'nearly' converge on the retina. But the convergence is not perfect.
• So the image formed on the retina will not be perfect. It will be 'blurred'.
• So a person with such defects will see only a blurred image
4. What is the remedy for this defect?
Consider fig.7.29(b). After passing through the lens, two sets of rays are shown
(i) A set consisting of two white rays
♦ This is the same shown in fig.a
(ii) A set consisting of two green rays
• If we follow the white set, we can see that the problem will not be remedied.
♦ They intersect behind the retina
• If we follow the green set, we can see that the problem will be remedied.
♦ They intersect exactly on the retina
5. So we want the green set. That is., after passing through the lens, the rays must follow the green lines. How can we make that happen?
• We notice that, the green rays 'converge more' than the white rays. That is why they are able to intersect exactly on the retina
• So if we can make the rays to 'converge more', the defect will be remedied
6. How can we make them to 'converge more'?
Ans: By using a converging lens.
A convex lens is a converging lens. So we can use it. It is shown in fig.7.29(c)
7. So we must put a convex lens in front of the eye to remedy this defect
• The convex lens in front of the eye, will create some degree of convergence even before the light enters the eye
• Then the lens in the eye will need to do only a 'lesser work' for making the rays intersect exactly on the retina
• The power of the 'convex lens to be used' will be prescribed by the doctor after examination
Case 2:
Consider fig.7.30 below. We will do an analysis of the fig.:
 |
| Fig.7.30 |
• But the point of convergence is not on the retina
• The point of convergence is in front of the retina. What could be the reason for this?
2. It may be due to any one or both of the two reasons given below:
(i) The lens has increased power
• 'Increase in the power' of lens means that the lens 'cause a bending, which is more than what is required'.
(ii) The eyeball is longer than normal so that the retina moved backwards
3. What will be the result when such defects occur?
• We see that a perfect image will be formed in front of the retina.
• But that image has no use. The retina can send signals based on only those images which fall exactly on it
■ We see that rays 'nearly' converge on the retina. But the convergence is not perfect.
• So the image formed on the retina will not be perfect. It will be 'blurred'.
• So a person with such defects will see only a blurred image
4. What is the remedy for this defect?
Consider fig.7.30(b). After passing through the lens, two sets of rays are shown
(i) A set consisting of two white rays
♦ This is the same shown in fig.a
(ii) A set consisting of two green rays
• If we follow the white set, we can see that the problem will not be remedied.
♦ They intersect in front of the retina
• If we follow the green set, we can see that the problem will be remedied.
♦ They intersect exactly on the retina
5. So we want the green set. That is., after passing through the lens, the rays must follow the green lines. How can we make that happen?
• We notice that, the green rays 'converge less' than the white rays. That is why they are able to intersect exactly on the retina
• So if we can make the rays to 'converge less', the defect will be remedied
6. How can we make them to 'converge less'?
Ans: 'Converging less' is equivalent to 'diverging more'
• So, by using a diverging lens, we can remedy this defect.
• A concave lens is a diverging lens. So we can use it
7. So we must put a concave lens in front of the eye to remedy this defect
• The concave lens in front of the eye, will create some degree of divergence even before the light enters the eye
• Then the rays will intersect exactly on the retina, even if the lens of the eye has increased power
• The power of the 'concave lens to be used' will be prescribed by the doctor after examination
The two defects that we saw above are given special names:
Case 1:
This is called Hypermetropia or Far-sightedness
The definition is:
■ Since the image is formed behind the retina, instead of being formed at the retina, even though distant objects are clearly seen, nearer objects cannot be seen. This defect of the eye is called Hypermetropia or Far-sightedness
• In the above definition, we see a peculiarity:
• It is said that, 'distant objects can be clearly seen'. That means people with hypermetropia:
♦ Can see distant objects
♦ Cannot see near objects
• How is that possible?
We will write the answer in steps:
1. Consider the fig.7.31(a) below:
 |
| Fig.7.31 |
• The white rays from the near object O are neatly converging on the retina
2. Another set of green rays are shown in the same fig.a
• Those rays are shown to be parallel to each other.
3. Why are they parallel to each other?
• The answer is that, those rays come from distant objects, and such rays will be always parallel to each other
4. In fig.7.31(b), two additional arcs are shown
• One magenta arc and one yellow arc
• The magenta arc shows the angle between incident white ray and refracted white ray
• The yellow arc shows the angle between incident green ray and refracted white ray
• Obviously, the angle indicated by the magenta arc is larger than that indicated by the yellow arc
5. So now compare the incident green set and the incident white set in fig.b
• We can see that, 'bending through a greater angle' is required to bring the white rays together on to the retina
• In other words, when compared to the white rays, a 'lesser effort' by the lens is sufficient to bring the green rays together on to the retina
6. Since 'lesser effort' is sufficient, even eyes with hypermetropia can make the rays from distant objects to intersect on the retina
• In other words, persons with hypermetropia will be able to see distant objects with out external help.
• While they cannot see near objects clearly. So another name for this defect is Far-sightedness.
Case 2:
This is called Myopia or Near-sightedness
The definition is:
■ Since the image is formed in front of the retina, instead of being formed at the retina, even though nearby objects are clearly seen, distant objects cannot be seen. This defect of the eye is called Myopia or Near-sightedness
• In the above definition, we see a peculiarity:
It is said that, 'near objects can be clearly seen'. That means people with myopia:
♦ Can see near objects
♦ Cannot see distant objects
• How is that possible?
The answer is just opposite to the one that we saw for case 1. But we will write the steps again:
1. Consider the same fig.7.31 above.
• It shows an eye with out any defects.
• The white rays from the near object O are neatly converging on the retina
2. Another set of green rays are shown in the same fig.a
• Those rays are shown to be parallel to each other.
3. Why are they parallel to each other?
• The answer is that, those rays come from distant objects, and such rays will be always parallel to each other
4. In fig.7.31(b), two additional arcs are shown
• One magenta arc and one yellow arc
• The magenta arc shows the angle between incident white ray and refracted white ray
• The yellow arc shows the angle between incident green ray and refracted white ray
• Obviously, the angle indicated by the magenta arc is larger than that indicated by the yellow arc
5. So now compare the green set and the white set.
• We can see that, 'more bending' is required to bring the white rays together on to the retina
• In other words, when compared to the white rays, a 'lesser effort' by the lens is sufficient to bring the green rays together on to the retina
6. Since 'lesser effort' is sufficient, the eyes with myopia (which has increased converging power) will be converging the green rays before they reach the retina.
• Converging of rays before reaching the retina is not useful
7. The white rays require 'more bending'. So the extra converging power available will be compensated.
• That is., the white rays will converge on the retina. Which means that near objects can be clearly seen without external help. So another name for this defect is Near-sightedness.
Astigmatism
1. Normally, the lens of the eye is smoothly curved in all directions.2. But if the curvature is not smooth, the surface of the lens will be irregular.
3. In such a situation, the light rays from both near and far objects will not be refracted properly.
4. They will not intersect on the retina. So both near and far objects will appear to be blurred.
5. This condition is called astigmatism.
6. This can be remedied by using cylindrical lens of appropriate power.
Presbyopia
1. We have seen that, the rays from near objects need 'greater bending'.2. For this, the power of the lens of the eye will have to be increased.
• This is achieved with the help of the ciliary muscles
3. But in elderly people, the ciliary muscles may not be able to provide the necessary curvature to the lens.
4. So elderly people will have to hold the objects at a distance greater than 25 cm, to see them properly. This condition is called presbyopia.
5. This can be remedied by using convex lens of appropriate power
Doctor's prescription
• The prescription made by the doctor will show the power of the required lens for rectifying the defects of the eye.
• Power of a lens is related to it's focal length
• Let us see an example:
1. Focal length of a lens is 20 cm
2. Express this focal length in metres:
• We have: 20 cm = 0.20 m
3. Take the reciprocal
• We have: Reciprocal of 0.20 = 1⁄0.02 = 100⁄20 = 5
4. This '5' is the power of the lens.
We can write the definition of power:
■ Power of a lens is the reciprocal of focal length expressed in metres.
• Power = 1⁄f
♦ Where f is the focal length expressed in metres
• Unit of power is dioptre. It is represented by D
■ Power of convex lens is positive and that of a concave lens is negative
But why do we take the reciprocal of the focal length?
The answer can be written in steps:
1. Consider fig.7.13 that we saw in a previous section. For convenience, it is shown again below:
• If the convex lens has greater ability to converge rays, we can say that it has greater power
• If the convex lens has lesser ability to converge rays, we can say that it has lesser power
3. But from the fig.a, it is clear that:
• If greater convergence is achieved, the focus F will move towards the lens
♦ That is., focal length will decrease
• If lesser convergence is achieved, the focus F will move away from the lens
♦ That is., focal length will increase
4. So we can conclude. In the case of a convex lens:
• Greater power means lesser focal length
• Lesser power means greater focal length
5. So there is an 'inverse relation'
• Such a relation can be expressed only if we put 'f' in the denominator.
• Because in the relation Power = 1⁄f,
♦ When f increases, power decreases
♦ When f decreases, power increases
■ So we indeed have to take the reciprocal of 'f'
6. Now consider the concave lens in fig.b. It is clear that concave lens is used for diverging light rays
• If the concave lens has greater ability to diverge rays, we can say that it has greater power
• If the concave lens has lesser ability to diverge rays, we can say that it has lesser power
7. But from the fig.b, it is clear that:
• If greater divergence is achieved, the focus F will move towards the lens
♦ That is., focal length will decrease
• If lesser divergence is achieved, the focus F will move away from the lens
♦ That is., focal length will increase
8. So we can conclude. In the case of a concave lens:
• Greater power means lesser focal length
• Lesser power means greater focal length
9. So there is an 'inverse relation'
• Such a relation can be expressed only if we put 'f' in the denominator.
• Because in the relation Power = 1⁄f,
♦ When f increases, power decreases
♦ When f decreases, power increases
■ So for the concave lens also, we indeed have to take the reciprocal of 'f'
Solved example 7.6
The powers of the lenses in the prescription of the doctor are +1.50 D and +1 D. What are the focal lengths of these two lenses? What type of lenses are they?
Solution:
Case 1:
1. We have: Power = +1.5 D = 1⁄f
♦ Where f is the focal length expressed in metres
2. So f = 1⁄+1.5 = +10⁄15 = +2⁄3 = +0.666 metres = +66.6 cm
3. Since the focal length is positive, it is a convex lens
Case 2:
1. We have: Power = +1 D = 1⁄f
♦ Where f is the focal length expressed in metres
2. So f = 1⁄+1 = +1 metre = +100 cm
3. Since the focal length is positive, it is a convex lens
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