Derivation of the Lens formula

Derivation of lens formula:
■ In section 7.4, we saw the application of the lens equation: ¹⁄_f = ¹⁄_v - ¹⁄_u  
In this section we will derive it. The steps are written below:
1. In the figure below, 
• P is the optic centre of a convex lens
• OB is the object
• IM is the image

• The two foci are given distinct names F₁ and F₂
   ♦ Distinct names are given for 'convenience in writing the steps' only
   ♦ Length PF₁ is equal to PF₂
• The ray parallel to the principal axis meets the y axis at C
2. Consider the two right triangles: ⊿POB and ⊿PMI:
(i) ∠BPO = ∠IPM (∵ they are opposite angles)
    ♦ This is indicated by the 'single green arcs' on opposite sides of P
(ii) ∠BOP = ∠IMP (∵ they are both right angles)
• The OB is kept perpendicular to the principal axis. So IM will be formed perpendicular to the y axis
• Thus ∠BOP and ∠IMP are right angles 
3. Since two angles are equal, the third angle will also be equal
So we get: ∠OBP = ∠MIP
    ♦ This is indicated by the 'double green arcs' at B and I
4. Thus all the three angles in both the triangles are the same
■ So the two triangles are similar
5. Now we apply a special property that is applicable to any two similar triangles (Details here):

^{side opposite smallest angle in ΔPOB}⁄_{side opposite smallest angle in ΔPMI}

= ^{side opposite medium angle in ΔPOB}⁄_{side opposite medium angle in ΔPMI}

= ^{side opposite largest angle in ΔPOB}⁄_{side opposite largest angle in ΔPMI}

6. But angles in the two triangles are the same. That is.,

• Smallest angle in ΔPOB = Smallest angle in ΔPMI

• Medium angle in ΔPOB = Medium angle in ΔPMI

• Largest angle in ΔPOB = Largest angle in ΔPMI
7. So we can write this:

• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,

^{side opposite P in ΔPOB}⁄_{side opposite P in ΔPMI}
= ^{side opposite B in ΔPOB}⁄_{side opposite I in ΔPMI}
= ^{side opposite O in ΔPOB}⁄_{side opposite M in ΔPMI}
8. So we get: ^OB⁄_IM = ^PO⁄_PM = ^PB⁄_PI 
• But OB = PC. So the first fraction in the above equation can be changed. We get:
^PC⁄_IM = ^PO⁄_PM = ^PB⁄_PI

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9. There is one more pair of similar triangles in the above fig.
Consider the two right triangles: ⊿F₂PC and ⊿F₂MI:
(i) ∠PF₂C = ∠MF₂I (∵ they are opposite angles)
(ii) ∠CPF₂ = ∠IMF₂ (∵ they are both right angles)
10. Since two angles are equal, the third angle will also be equal
So we get: ∠PCF₂ = ∠MIF₂
11. Thus all the three angles in both the triangles are the same
■ So the two triangles are similar
12. We can apply the same procedure that we applied above. We will write the steps again:

^{side opposite smallest angle in ΔF₂PC}⁄_{side opposite smallest angle in ΔF2MI}

= ^{side opposite medium angle in ΔF₂PC}⁄_{side opposite medium angle in ΔF2MI}

= ^{side opposite largest angle in ΔF₂PC}⁄_{side opposite largest angle in ΔF2MI}

13. But angles in the two triangles are the same. That is.,

• Smallest angle in ΔF₂PC = Smallest angle in ΔF₂MI

• Medium angle in ΔF₂PC = Medium angle in ΔF₂MI

• Largest angle in ΔF₂PC = Largest angle in ΔF₂MI
14. So we can write this:

• Ratio of the sides opposite equal angles in the two similar triangles are the same. That is.,

^{side opposite F₂ in ΔF₂PC}⁄_{side opposite F2 in ΔF2MI}

= ^{side opposite C in ΔF₂PC}⁄_{side opposite I in ΔF2MI}

= ^{side opposite P in ΔF₂PC}⁄_{side opposite M in ΔF2MI}

15. So we get: ^PC⁄_IM = ^PF₂⁄_MF2= ^CF₂⁄_IF2
Compare the above result in (15) with the result in (8):
16. We see that, the first fraction is the same in both results
• So the 'second and third fractions in (8)' can be equated to 'second and third fractions in (15)'
• We get: ^PO⁄_PM = ^PB⁄_PI = ^PF₂⁄_MF2= ^CF₂⁄_IF2
17. For our present derivation, we need only the horizontal distances
• We do not want the vertical and sloping distances
    ♦ Consider the first fraction in (16). PO and PM are horizontal distances. So we keep this fraction 
    ♦ Consider the second fraction in (16). PB and PI are sloping distances. So we discard this fraction 
    ♦ Consider the third fraction in (16). PF₂ and MF₂ are horizontal distances. So we keep this fraction
    ♦ Consider the fourth fraction in (16). CF₂ and IF₂ are sloping distances. So we discard this fraction 
18. Thus the required result is:
^PO⁄_PM = ^PF₂⁄_MF2
19. But 
• PO = -u (negative sign is given because PO is measured along the negative x direction)
• PM = v
• PF₂ = the focal length 'f'
• MF₂ = (PM - PF₂) = (v-f)
• So the result in (18) becomes:
^-u⁄_v = ^f⁄_(v-f).
⟹ -u(v-f) = vf ⟹ -uv + uf = vf
• Dividing both sides by uvf, we get:
^-1⁄_f + ¹⁄_v = ¹⁄_u.
⟹  ¹⁄_f = ¹⁄_v - ¹⁄_u 
• Thus the lens formula is derived

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Now we will prove the equation related to magnification. That is: m = ^h_i ⁄_ho = ^v⁄_u.
We will write the steps:
1. Consider the figure given at the beginning of this section.
• In the step (8) above, we proved that: ^OB⁄_IM = ^PO⁄_PM = ^PB⁄_PI
    ♦ Consider the third fraction. PB and PI are sloping distances. So we discard this fraction
2. Thus the required result is: ^OB⁄_IM = ^PO⁄_PM 
3. But
• OB = h_o 
• IM = h_i 
• PO = u
• PM = v
4. Substituting these values in (2), we get:  ^h_i ⁄_ho = ^v⁄_u 
• Thus the relation is proved

CONTENTS

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Chapter 7.4 - The Lens Formula and Magnification

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

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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

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■ 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

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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

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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} = ^h_i⁄_ho.
• So to find 'm':
    ♦ We need to measure the height of the object (h_o)
    ♦ We need to measure the height of the image (h_i)
• However, there is another method to calculate 'm'
• This is based on the fact that (^h_i⁄_ho) will be equal to (^v⁄_u)
• So we can write: m = ^h_i ⁄_ho = ^v⁄_u.
• Obviously it is easier to measure u and v. So 'm' can be calculated easily 
• The mathematical steps which prove that '(^h_i ⁄_ho) is equal to (^v ⁄_u)' can be seen here.

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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 = ^h_i ⁄_ho = ^h_i ⁄₃ =  ⁶⁰⁄_-30 = -2
4. Equating the results in (2) and (3), we get: ^h_i ⁄₃ = -2
⟹ h_i  = -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: h_i = h_o × ¹⁄₂
• So m = ^h_i ⁄_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 = ^h_i ⁄_ho = ^v⁄_u = ³⁰⁄_-15 = -2
⟹ h_i = -2 × h_o = 2 × 3 = -6 cm
• The negative height shows that the image is inverted 

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In the next section, we will see Eye and vision.

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