In the previous section we saw that the ray diagrams when the object is placed at different positions in front of the lens. In this section, we will see an experiment based on those ray diagrams.
Before doing the experiment, we must do an analysis of the fig.7.26 below.
We will write the analysis in steps:
1. In fig.7.26(a), the object is placed at 6.25 cm to the left of P. So u = -6.25 cm
2. The two rays intersect at 11.11 cm to the right of P. So v = 11.11 cm
3. If we place a screen at this 11.11 cm, we will get a clear image of the object
4. Now consider fig.b
• The lens is kept at the same position.
• The screen is kept at the same position
• But the object is moved.
5. The image disappears from the screen. Why does it disappear?
• The reason is that, the two rays no longer intersect at 11.11 cm.
• When the object was moved, the point of intersection of the two rays also moved
• So we will not get an image at 11.11 cm
6. Why does the point of intersection move?
The answer can be written in steps:
(i) The top most ray parallel to the principal axis is the same in both the figs
(ii) But consider the other ray passing through P:
♦ In fig.a, it passes through (-6.25, 2.25) and P(0,0)
♦ In fig.b, it passes through (-7.25, 2.25) and P(0,0)
♦ [The y coordinate 2.25 is the height of the object]
• So this ray is not the same in figs (a) and (b)
• Thus the point of intersection will differ
■ Now, to make the image appear again, we will have to move the screen
7. Based on the above analysis, we can write:
■ If the object is moved keeping the lens at the same position, the screen will also have to be moved to get the image
Now we can begin the experiment:
1. Put a lighted candle in front of a convex lens
2. Put a screen on the other side of the lens
• Adjust the position of the screen to get a clear image of the candle
3. Note down u and v
• They are recorded as -30 and 60 respectively in the first row of the table below:
4. With out changing the position of the lens, move the candle.
• Let it be moved by 10 cm to towards the left. Then new u = -40 cm
5. Adjust the position of the screen to obtain a clear image of the candle
• New v is obtained as 40 cm. These are recorded in the second row
6. Repeat the trial.
• u changes to -50 and v changes to 33.33
7. The observations are over. Now we do some calculations:
• Fill up the fourth column with '1⁄u' values
• Fill up the fifth column with '1⁄v' values
8. Subtract '1⁄u' from '1⁄v' and fill up the sixth column
• We see that the heading of this sixth column is '[1⁄f = 1⁄v - 1⁄u]'
• It is an equation. According to this equation:
• When we Subtract '1⁄u' from '1⁄v', we get the reciprocal of the focal length 'f'
9. So from the first trial we get: Reciprocal of 'f' is 0.05
♦ That is., 1⁄f = 0.05 ⟹ f = 1⁄0.05 = 20
• From the second trial we get: Reciprocal of 'f' is 0.05
♦ That is., 1⁄f = 0.05 ⟹ f = 1⁄0.05 = 20
• From the third trial we get: Reciprocal of 'f' is 0.050003
♦ That is., 1⁄f = 0.050003 ⟹ f = 1⁄0.050003 = 19.9988
10. Note that, 0.05 and 0.050003 are nearly equal
• Their reciprocals will also be nearly equal
• We find that '20' and '19.9988' are indeed nearly equal
11. The equation '[1⁄f = 1⁄v - 1⁄u]' can be easily simplified to the form: '[f = uv⁄(u-v) ]'
• Using this, we can directly calculate 'f'. The fourth, fifth and sixth columns can be avoided.
• The results obtained in this way are recorded in the seventh column
• Note that, we get the same values for 'f'
12. Which ever be the method, we must calculate the average of the results from all the trials;
• Trial 1 gives: f = 20
• Trial 2 gives: f = 20
• Trial 3 gives: f = 19.9988
■ So average f = (20+20+19.9988)⁄3 = 19.9996 ≈ 20 cm
■ The equation '[1⁄f = 1⁄v - 1⁄u]' that we saw above is called the lens equation.
It's derivation can be seen here.
• Simplifying it, we got: [f = uv⁄(u-v) ]
• Two more forms can be obtained:
[v = uf⁄(u+f)] and [u = fv⁄(f-v)]
• These equations can be used for solving problems
Solved example 7.1
When an object is placed at a distance of 15 cm from a convex lens, a real image is formed at a distance of 30 cm. What is the focal length of the lens?
Solution:
1. Given that, the object and image is real.
• For a convex lens, object and it's real image will always be on opposite sides
• So one distance will be negative and the other, positive.
2. We will assume that object is on the left side and image is on the right side of the lens. Thus we can write:
u = -15 cm, v = 30 cm
3. We have: f = uv⁄(u-v)
Substituting the values, we get: f = (-15×30)⁄(-15-30) = -450⁄-45 = 10 cm
4. Note that we got a positive value for 'f'
• The focal length of convex lens is always positive
• The focal length of concave lens is always negative
Solved example 7.2
The focal length of a concave lens is 20 cm. If an object is kept at a distance of 30 cm from the lens, find out the distance of the image formed
Solution:
1. For a concave lens, both object and image will always be on the same side of the lens. We will assume that, they are both on the left side of the lens.
• Then we can write: u = -30
2. The focus of a concave lens is virtual. So it's focal length is negative.
• Then we can write: f = -20 cm
3. We have: v = uf⁄(u+f)
• Substituting the values, we get: v = (-30×-20)⁄(-30-20) = +600⁄-50 = -12 cm
4. Note that a negative value is obtained for v
• u was also given a negative value
• So both the object and image are indeed on the same side (left) of the lens
(i) The image may be enlarged than the object
(ii) The image may be of the same size as the object
(iii) The image may be diminished than the object
• Consider the case 1. We would want to know 'how much enlargement' occurred?
We will write the steps:
1. Both height and width will be enlarged by the same factor. Let 'm' be the factor
2. Then we can say: [Height of object] × m = [Height of image]
• Rearranging this equation, we get: m = Height of image⁄Height of object
3. This gives us an easy method to find 'm'
• It is clear that,
(i) If 'm' is greater than 1, the image will be enlarged
(ii) If 'm' is equal to 1, the image will be of the same size
(iii) If 'm' is less than 1, the image will be diminished
4. This 'm' is called magnification.
■ We can write the definition:
Magnification is the ratio of the height of the image to the height of the object. It shows how many times the image is as large as the object
5. Magnification = Height of image⁄Height of object = hi⁄ho.
• So to find 'm':
♦ We need to measure the height of the object (ho)
♦ We need to measure the height of the image (hi)
• However, there is another method to calculate 'm'
• This is based on the fact that (hi⁄ho) will be equal to (v⁄u)
• So we can write: m = hi ⁄ho = v⁄u.
• Obviously it is easier to measure u and v. So 'm' can be calculated easily
• The mathematical steps which prove that '(hi ⁄ho) is equal to (v ⁄u)' can be seen here.
Solved example 7.3
When an object of height 3 cm is placed at a distance of 30 cm from a lens, a real image is formed at a distance of 60 cm. Find out the height of the image
Solution:
1. Given that, the image is real. So the object and image are on the opposite sides of the lens
• Let the object be on the left side. Then u = -30 cm
• Let the image be on the right side. Then v = 60 cm
2. We have: magnification = v⁄u = 60⁄-30 = -2
3. We also have: magnification = hi ⁄ho = hi ⁄3 = 60⁄-30 = -2
4. Equating the results in (2) and (3), we get: hi ⁄3 = -2
⟹ hi = -6 cm
• The negative sign indicates that the height of the image is measured below the principal axis. In other words, the image is formed below the principal axis
Solved example 7.4
A concave lens of focal length 10 cm produces an image which is half the size of the object. How far away is the object from the lens? Find also the position and nature of the image.
Solution:
1. Focal length of a concave lens should be taken as negative. So we can write: f = -10 cm
2. Given that image is half the size of the object. So we can write: hi = ho × 1⁄2
• So m = hi ⁄ho = 1⁄2
3. We also have: magnification = v⁄u
4. Equating the results in (2) and (3), we get: v⁄u = 1⁄2
⟹ u = 2v
5. We have: 1⁄f = 1⁄v - 1⁄u
Substituting u = 2v and f = -10, we get:
1⁄-10 = 1⁄v - 1⁄2v
⟹ 1⁄-10 = (2-1)⁄2v = 1⁄2v =
⟹ 2v = -10 ⟹ v = -5 cm
6. Then u = 2v = -10 cm
7. Both u and v are negative. So both are on the left side of the lens
• The object is 10 cm from optic centre P
• The image is 5 cm from optic centre P
• For a concave lens, image will always be virtual
Solved example 7.5
An object of height 3 cm is placed in front of a convex lens of focal length 10 cm at a distance of 15 cm.
(a) What is the distance of the image formed?
(b) What is the nature of the image?
(c) What is the height of the image?
Solution:
Part (a): We have to find v
We have: v = uf⁄(u+f)
• Substituting the values, we get: v = (-15×10)⁄(-15+10) = -150⁄-5 = 30 cm
Part (b):
1. We obtained a positive value for v. The distance u was taken as negative
• That means, image is formed on the other side of the lens
• The image formed by a convex lens in this manner will be always real and inverted
Part (c):
• We have: magnification m = hi ⁄ho = v⁄u = 30⁄-15 = -2
⟹ hi = -2 × ho = 2 × 3 = -6 cm
• The negative height shows that the image is inverted
Before doing the experiment, we must do an analysis of the fig.7.26 below.
We will write the analysis in steps:
1. In fig.7.26(a), the object is placed at 6.25 cm to the left of P. So u = -6.25 cm
2. The two rays intersect at 11.11 cm to the right of P. So v = 11.11 cm
3. If we place a screen at this 11.11 cm, we will get a clear image of the object
4. Now consider fig.b
• The lens is kept at the same position.
• The screen is kept at the same position
• But the object is moved.
5. The image disappears from the screen. Why does it disappear?
• The reason is that, the two rays no longer intersect at 11.11 cm.
• When the object was moved, the point of intersection of the two rays also moved
• So we will not get an image at 11.11 cm
6. Why does the point of intersection move?
The answer can be written in steps:
(i) The top most ray parallel to the principal axis is the same in both the figs
(ii) But consider the other ray passing through P:
♦ In fig.a, it passes through (-6.25, 2.25) and P(0,0)
♦ In fig.b, it passes through (-7.25, 2.25) and P(0,0)
♦ [The y coordinate 2.25 is the height of the object]
• So this ray is not the same in figs (a) and (b)
• Thus the point of intersection will differ
■ Now, to make the image appear again, we will have to move the screen
7. Based on the above analysis, we can write:
■ If the object is moved keeping the lens at the same position, the screen will also have to be moved to get the image
Now we can begin the experiment:
1. Put a lighted candle in front of a convex lens
2. Put a screen on the other side of the lens
• Adjust the position of the screen to get a clear image of the candle
3. Note down u and v
• They are recorded as -30 and 60 respectively in the first row of the table below:
• Let it be moved by 10 cm to towards the left. Then new u = -40 cm
5. Adjust the position of the screen to obtain a clear image of the candle
• New v is obtained as 40 cm. These are recorded in the second row
6. Repeat the trial.
• u changes to -50 and v changes to 33.33
7. The observations are over. Now we do some calculations:
• Fill up the fourth column with '1⁄u' values
• Fill up the fifth column with '1⁄v' values
8. Subtract '1⁄u' from '1⁄v' and fill up the sixth column
• We see that the heading of this sixth column is '[1⁄f = 1⁄v - 1⁄u]'
• It is an equation. According to this equation:
• When we Subtract '1⁄u' from '1⁄v', we get the reciprocal of the focal length 'f'
9. So from the first trial we get: Reciprocal of 'f' is 0.05
♦ That is., 1⁄f = 0.05 ⟹ f = 1⁄0.05 = 20
• From the second trial we get: Reciprocal of 'f' is 0.05
♦ That is., 1⁄f = 0.05 ⟹ f = 1⁄0.05 = 20
• From the third trial we get: Reciprocal of 'f' is 0.050003
♦ That is., 1⁄f = 0.050003 ⟹ f = 1⁄0.050003 = 19.9988
10. Note that, 0.05 and 0.050003 are nearly equal
• Their reciprocals will also be nearly equal
• We find that '20' and '19.9988' are indeed nearly equal
11. The equation '[1⁄f = 1⁄v - 1⁄u]' can be easily simplified to the form: '[f = uv⁄(u-v) ]'
• Using this, we can directly calculate 'f'. The fourth, fifth and sixth columns can be avoided.
• The results obtained in this way are recorded in the seventh column
• Note that, we get the same values for 'f'
12. Which ever be the method, we must calculate the average of the results from all the trials;
• Trial 1 gives: f = 20
• Trial 2 gives: f = 20
• Trial 3 gives: f = 19.9988
■ So average f = (20+20+19.9988)⁄3 = 19.9996 ≈ 20 cm
■ The equation '[1⁄f = 1⁄v - 1⁄u]' that we saw above is called the lens equation.
It's derivation can be seen here.
• Simplifying it, we got: [f = uv⁄(u-v) ]
• Two more forms can be obtained:
[v = uf⁄(u+f)] and [u = fv⁄(f-v)]
• These equations can be used for solving problems
Solved example 7.1
When an object is placed at a distance of 15 cm from a convex lens, a real image is formed at a distance of 30 cm. What is the focal length of the lens?
Solution:
1. Given that, the object and image is real.
• For a convex lens, object and it's real image will always be on opposite sides
• So one distance will be negative and the other, positive.
2. We will assume that object is on the left side and image is on the right side of the lens. Thus we can write:
u = -15 cm, v = 30 cm
3. We have: f = uv⁄(u-v)
Substituting the values, we get: f = (-15×30)⁄(-15-30) = -450⁄-45 = 10 cm
4. Note that we got a positive value for 'f'
• The focal length of convex lens is always positive
• The focal length of concave lens is always negative
Solved example 7.2
The focal length of a concave lens is 20 cm. If an object is kept at a distance of 30 cm from the lens, find out the distance of the image formed
Solution:
1. For a concave lens, both object and image will always be on the same side of the lens. We will assume that, they are both on the left side of the lens.
• Then we can write: u = -30
2. The focus of a concave lens is virtual. So it's focal length is negative.
• Then we can write: f = -20 cm
3. We have: v = uf⁄(u+f)
• Substituting the values, we get: v = (-30×-20)⁄(-30-20) = +600⁄-50 = -12 cm
4. Note that a negative value is obtained for v
• u was also given a negative value
• So both the object and image are indeed on the same side (left) of the lens
Magnification
• When we discussed ray diagrams, we saw the following:(i) The image may be enlarged than the object
(ii) The image may be of the same size as the object
(iii) The image may be diminished than the object
• Consider the case 1. We would want to know 'how much enlargement' occurred?
We will write the steps:
1. Both height and width will be enlarged by the same factor. Let 'm' be the factor
2. Then we can say: [Height of object] × m = [Height of image]
• Rearranging this equation, we get: m = Height of image⁄Height of object
3. This gives us an easy method to find 'm'
• It is clear that,
(i) If 'm' is greater than 1, the image will be enlarged
(ii) If 'm' is equal to 1, the image will be of the same size
(iii) If 'm' is less than 1, the image will be diminished
4. This 'm' is called magnification.
■ We can write the definition:
Magnification is the ratio of the height of the image to the height of the object. It shows how many times the image is as large as the object
5. Magnification = Height of image⁄Height of object = hi⁄ho.
• So to find 'm':
♦ We need to measure the height of the object (ho)
♦ We need to measure the height of the image (hi)
• However, there is another method to calculate 'm'
• This is based on the fact that (hi⁄ho) will be equal to (v⁄u)
• So we can write: m = hi ⁄ho = v⁄u.
• Obviously it is easier to measure u and v. So 'm' can be calculated easily
• The mathematical steps which prove that '(hi ⁄ho) is equal to (v ⁄u)' can be seen here.
Solved example 7.3
When an object of height 3 cm is placed at a distance of 30 cm from a lens, a real image is formed at a distance of 60 cm. Find out the height of the image
Solution:
1. Given that, the image is real. So the object and image are on the opposite sides of the lens
• Let the object be on the left side. Then u = -30 cm
• Let the image be on the right side. Then v = 60 cm
2. We have: magnification = v⁄u = 60⁄-30 = -2
3. We also have: magnification = hi ⁄ho = hi ⁄3 = 60⁄-30 = -2
4. Equating the results in (2) and (3), we get: hi ⁄3 = -2
⟹ hi = -6 cm
• The negative sign indicates that the height of the image is measured below the principal axis. In other words, the image is formed below the principal axis
Solved example 7.4
A concave lens of focal length 10 cm produces an image which is half the size of the object. How far away is the object from the lens? Find also the position and nature of the image.
Solution:
1. Focal length of a concave lens should be taken as negative. So we can write: f = -10 cm
2. Given that image is half the size of the object. So we can write: hi = ho × 1⁄2
• So m = hi ⁄ho = 1⁄2
3. We also have: magnification = v⁄u
4. Equating the results in (2) and (3), we get: v⁄u = 1⁄2
⟹ u = 2v
5. We have: 1⁄f = 1⁄v - 1⁄u
Substituting u = 2v and f = -10, we get:
1⁄-10 = 1⁄v - 1⁄2v
⟹ 1⁄-10 = (2-1)⁄2v = 1⁄2v =
⟹ 2v = -10 ⟹ v = -5 cm
6. Then u = 2v = -10 cm
7. Both u and v are negative. So both are on the left side of the lens
• The object is 10 cm from optic centre P
• The image is 5 cm from optic centre P
• For a concave lens, image will always be virtual
Solved example 7.5
An object of height 3 cm is placed in front of a convex lens of focal length 10 cm at a distance of 15 cm.
(a) What is the distance of the image formed?
(b) What is the nature of the image?
(c) What is the height of the image?
Solution:
Part (a): We have to find v
We have: v = uf⁄(u+f)
• Substituting the values, we get: v = (-15×10)⁄(-15+10) = -150⁄-5 = 30 cm
Part (b):
1. We obtained a positive value for v. The distance u was taken as negative
• That means, image is formed on the other side of the lens
• The image formed by a convex lens in this manner will be always real and inverted
Part (c):
• We have: magnification m = hi ⁄ho = v⁄u = 30⁄-15 = -2
⟹ hi = -2 × ho = 2 × 3 = -6 cm
• The negative height shows that the image is inverted
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