In the previous section we saw that the rays emerging from the object, passes through the lens and then converge to give points on the image. To draw the exact paths taken by the rays, we must be familiar with some rules. They are given below:
Rule 1:
• When a light ray passes through the optic centre of a thin lens, it does not undergo any deviation. This is shown in figs.7.16 (a) and (b) below:
• We can see that the rays are passing through P. The lens is not capable of causing any deviation to such rays
Rule 2:
A ray of light falling parallel to the principal axis of a convex lens passes through the principal focus after refraction. This is shown in fig.7.17 below:
Rule 3:
A ray of light falling parallel to the principal axis of a concave lens appears to emerge from the focus on the same side of the lens. This is shown in fig.7.18 below:
• Note that, after refraction, the ray diverges away. We can only trace it's path backwards by drawing a dashed line on paper. When we draw such a dashed line, we will reach the focus F
Rule 4:
A ray of light passing through the principal focus of a convex lens passes parallel to the principal axis after refraction. This is shown in fig.7.19 below:
This can be explained using the following steps:
1. Light rays emerge in all directions from an object
2. Among those rays in (1) abive, some will pass through F
3. Among those rays in (2) above, some will fall on the lens
4. All rays in (3) will undergo deviation.
• After the deviation, they will become parallel to the principal axis
• This can be considered as the reverse of Rule 2
Rule 5:
• All the light rays coming from an 'object at infinity' (or a 'distant object') are considered to be parallel to each other.
• We have seen that such parallel rays will converge at F. See fig.7.13 of the previous section
Now we are ready to draw ray diagrams. The ray diagrams will help us to predict three items:
(i) Position of the image
(ii) Nature of the image
(iii) Size of the image
Case 1: Object beyond 2F
This is shown in fig.7.20 below:
It is convenient to draw the ray diagram on a graph paper. Let us write the steps:
1. The optic centre P must be placed at the 'origin of the graph'
• 'Origin of the graph' is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 9.5 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the object from P is denoted as 'u'
■ So we can write: u = -9.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 6.9 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = +6.9 cm or simply 7 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is +2.25 cm or simply 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 1.63 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -1.63 cm
10. Considering the numeric values only, we have:
• Height of the object = 2.25 cm
• Height of the image = 1.63 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed beyond 2F:
(i) Position of the image: Between F and 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Diminished
Case 2: Object at 2F
This is shown in fig.7.21 below:
As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2F
■ So we can write: u = -8 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 8 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 8 cm
• Note that this is the distance of 2F (on the other side) from P
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is also 2.25 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -2.25 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = -2.25 cm
• So the height of the image has not changed.
• Not only the height, the thickness also will not change.
11. We can write the conclusion:
■ When the object is placed at 2F:
(i) Position of the image: At 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Same size
Case 3: Object between F and 2F
This is shown in fig.7.22 below:
As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 11.11 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 11.11 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 4 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -4 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 4 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and 2F:
(i) Position of the image: Beyond 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Enlarged
Case 4: Object at F
This is shown in fig.7.23 below:
As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at F. That is., 4 units to the left of P
■ So we can write: u = -4 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. After the refraction, the two rays become parallel to each other
• They will never meet
• Since they are parallel, they will not meet even if we draw them backwards
8. We can write the conclusion:
■ When the object is placed at F:
• Image is not formed
Case 4: Object between F and the lens
This is shown in fig.7.24 below:
As before, we will draw the ray diagram on a graph paper.
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2.5 units to the left of P
■ So we can write: u = -2.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
♦ This is shown by the dashed lines
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 6.66 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -6.66 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 6 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 6.0 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 6.0 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and lens:
(i) Position of the image: Same side of the object
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Enlarged
Case 5: Image formation by a concave lens
This is shown in fig.7.25 below:
As before, we will draw the ray diagram on a graph paper.
• The steps are as follows:
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. We will not be needing 2F for this case
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 3 that we saw above.
♦ After refraction, it will diverge away
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
♦ This is shown by the dashed line for ray (i)
♦ For ray (ii), the dashed line will fall on that ray itself. So it is not visible
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 2.21 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -2.21 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 0.79 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 0.79 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 0.79 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed anywhere in front of a concave lens:
(i) Position of the image: Same side of the object between F and the lens
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Diminished
Rule 1:
• When a light ray passes through the optic centre of a thin lens, it does not undergo any deviation. This is shown in figs.7.16 (a) and (b) below:
 |
| Fig.7.16 |
Rule 2:
A ray of light falling parallel to the principal axis of a convex lens passes through the principal focus after refraction. This is shown in fig.7.17 below:
 |
| Fig.7.17 |
A ray of light falling parallel to the principal axis of a concave lens appears to emerge from the focus on the same side of the lens. This is shown in fig.7.18 below:
 |
| Fig.7.18 |
Rule 4:
A ray of light passing through the principal focus of a convex lens passes parallel to the principal axis after refraction. This is shown in fig.7.19 below:
 |
| Fig.7.19 |
1. Light rays emerge in all directions from an object
2. Among those rays in (1) abive, some will pass through F
3. Among those rays in (2) above, some will fall on the lens
4. All rays in (3) will undergo deviation.
• After the deviation, they will become parallel to the principal axis
• This can be considered as the reverse of Rule 2
Rule 5:
• All the light rays coming from an 'object at infinity' (or a 'distant object') are considered to be parallel to each other.
• We have seen that such parallel rays will converge at F. See fig.7.13 of the previous section
Now we are ready to draw ray diagrams. The ray diagrams will help us to predict three items:
(i) Position of the image
(ii) Nature of the image
(iii) Size of the image
Case 1: Object beyond 2F
This is shown in fig.7.20 below:
 |
| Fig.7.20 |
1. The optic centre P must be placed at the 'origin of the graph'
• 'Origin of the graph' is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 9.5 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the object from P is denoted as 'u'
■ So we can write: u = -9.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 6.9 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = +6.9 cm or simply 7 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is +2.25 cm or simply 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 1.63 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -1.63 cm
10. Considering the numeric values only, we have:
• Height of the object = 2.25 cm
• Height of the image = 1.63 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed beyond 2F:
(i) Position of the image: Between F and 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Diminished
Case 2: Object at 2F
This is shown in fig.7.21 below:
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2F
■ So we can write: u = -8 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 8 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 8 cm
• Note that this is the distance of 2F (on the other side) from P
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is also 2.25 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -2.25 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = -2.25 cm
• So the height of the image has not changed.
• Not only the height, the thickness also will not change.
11. We can write the conclusion:
■ When the object is placed at 2F:
(i) Position of the image: At 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Same size
Case 3: Object between F and 2F
This is shown in fig.7.22 below:
 |
| Fig.7.22 |
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays meet at a point on the other side of the lens
• This meeting point is the image of the 'top of the object'
• From there we can draw the image upwards
8. The position of the image:
The image is formed at 11.11 units to the right of P
• We know that all horizontal measurements on the right of the y axis are considered as positive
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = 11.11 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 4 units
• We know that all vertical measurements below the x axis are considered as negative
■ So the height of the image in our present case is -4 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 4 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and 2F:
(i) Position of the image: Beyond 2F on the other side
(ii) Nature of the image: Real and inverted
(iii) Size of the image: Enlarged
Case 4: Object at F
This is shown in fig.7.23 below:
 |
| Fig.7.23 |
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at F. That is., 4 units to the left of P
■ So we can write: u = -4 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. After the refraction, the two rays become parallel to each other
• They will never meet
• Since they are parallel, they will not meet even if we draw them backwards
8. We can write the conclusion:
■ When the object is placed at F:
• Image is not formed
Case 4: Object between F and the lens
This is shown in fig.7.24 below:
 |
| Fig.7.24 |
• The steps are as follows (steps 1 to 4 are same as before. But we will write them again):
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. Then 2F will be '8 units' on either side of P
5. The object is placed at 2.5 units to the left of P
■ So we can write: u = -2.5 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 2 that we saw above.
♦ After refraction, it will pass through F
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
♦ This is shown by the dashed lines
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 6.66 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -6.66 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 6 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 6.0 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 6.0 cm
• So the height of the image has increased.
• Not only the height, the thickness also will increase. So a thicker yellow line and a bigger square are shown as the image
11. We can write the conclusion:
■ When the object is placed between F and lens:
(i) Position of the image: Same side of the object
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Enlarged
Case 5: Image formation by a concave lens
This is shown in fig.7.25 below:
 |
| Fig.7.25 |
• The steps are as follows:
1. The optic centre P must be placed at the origin of the graph
• Origin is at the point of intersection of x and y axes
2. The principal axis should coincide with the x axis
3. In our present case, the focal length is 4 cm.
• So F is marked '4 units' to the left of P on the x axis
• The other F is marked '4 units' to the right of P on the x axis
4. We will not be needing 2F for this case
5. The object is placed at 6.25 units to the left of P
■ So we can write: u = -6.25 cm
6. Now we can draw the rays:
• We know that, rays emerge from an object in all directions
• We consider two of them. Both emerging from the 'top of the object'
(i) The first one is parallel to the principal axis
♦ We know what will happen to such a ray. We know it from Rule 3 that we saw above.
♦ After refraction, it will diverge away
(ii) The second one passes through P
♦ We know what will happen to such a ray. We know it from Rule 1 that we saw above.
♦ It will not undergo any deviation
7. The two rays diverge from each other. They will not meet on the right side of the lens
• But if they are drawn backwards, they will meet at a point on the same side
♦ This is shown by the dashed line for ray (i)
♦ For ray (ii), the dashed line will fall on that ray itself. So it is not visible
• This meeting point is the image of the 'top of the object'
• From there we can draw the image downwards
8. The position of the image:
The image is formed at 2.21 units to the left of P
• We know that all horizontal measurements on the left of the y axis are considered as negative
• We have seen that the distance of the image from P is denoted as 'v'
■ So we can write: v = -2.21 cm
9. Now let us consider the vertical measurements:
There are two vertical measurements:
(i) Height of the object
• From the graph we can see that it is 2.25 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the object in our present case is 2.25 cm
(ii) Height of the image
• From the graph we can see that it is 0.79 units
• We know that all vertical measurements above the x axis are considered as positive
■ So the height of the image in our present case is 0.79 cm
10. We have:
• Height of the object = 2.25 cm
• Height of the image = 0.79 cm
• So the height of the image has decreased.
• Not only the height, the thickness also will decrease. So a thinner yellow line and a smaller square are shown as the image
11. We can write the conclusion:
■ When the object is placed anywhere in front of a concave lens:
(i) Position of the image: Same side of the object between F and the lens
(ii) Nature of the image: Virtual and erect
(iii) Size of the image: Diminished
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