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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