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 7.2 - Convex and Concave Lens

In the previous section we saw the details about refractive index. In this section we will learn about lens.

Lens is a transparent medium having spherical surfaces
From the above definition, we can say that, a lens should satisfy the following two conditions:
(i) It must be transparent
• That is., a lens can be made using glass, plastic or any similar materials. But it must be transparent
(ii) It should have spherical surfaces
• That is., any surface of a lens should be a part of a sphere. 
• It is just like saying: An arc is a part of a circle

Let us see the details about those spherical surfaces. We will write them in steps:
1. Consider the two spheres in fig.7.10 (a) below:
• They are identical spheres. That is., both have the same radii

Fig.7.10

• The centre of the left sphere is C₁ 
• The centre of the right sphere is C₂
• The red line joins C₁ and C₂ 
2. In fig.7.10(b), the two spheres are brought closer to each other.
• So close that, they intersect each other
• The distance C₁C₂ is now lesser than the C₁C₂ in fig.a.
3. The 'portion common to both the spheres' is retained as such
• The rest of the two spheres are erased off
• The result is shown in fig.c
• The retained portion in fig.c forms a convex lens
• Note that, the distance C₁C₂ is fib.c is the exact same as that in fig.b
4. The surfaces of the lens are thus parts of the surfaces of spheres

Now we will see the concave lens:
1. Consider the two spheres in fig.7.11 (a) below:
• They are identical spheres. That is., both have the same radii

Fig.7.11

• The centre of the left sphere is C₁ 
• The centre of the right sphere is C₂
• The red line joins C₁ and C₂ .
2. In the fig.a, the spheres are close to each other. But they are not touching each other
• In fig.b, the space between them is filled up with a transparent material
3. In fig.c, the two spheres are removed.
• The newly added transparent material is left behind. This forms a concave lens.
• Note that, the distance C₁C₂ is fib.c is the exact same as that in fig.b
4. The surfaces of the lens are thus parts of the surfaces of spheres

Let us see the properties of convex and concave lenses:
1. Optic centre
• Optic centre is the midpoint of a lens. 
• It is denoted by the letter 'P'. It is shown in fig.7.12 below:

Fig.7.12

2. Centre of curvature
• We have seen that each lens is associated with two spheres with centres C₁ and C₂
• Both C₁ and C₂ are eligible to be called as the 'centre of curvature' of the lens
• So the definition is:
Centre of curvature is the centre of the imaginary spheres of which the surfaces of the lens are parts. It is denoted by the letter 'C'
3. Principal axis
• The definition is:
Principal axis is the imaginary line that passes through the optic centre joining the two centres of curvatures
• So the principal axis should satisfy both the conditions given below:
(i) The principal axis must pass through both C₁ and C₂
(ii) The principal axis must pass through P
4. Principal focus
(a) Convex lens
• The definition is:
Light rays parallel and close to the principal axis converge at a point on the principal axis of a convex lens. This point is it's principal focus. It is denoted by the letter 'F'
• This will be clear from fig.7.13(a) below

Fig.7.13

• The red line is the principal axis. 
• Note that light rays will be travelling in all directions. In the fig.a above, we are considering only those which satisfies the following two conditions:
(i) Light rays must be parallel to the principal axis  
(ii) Light rays must be close to the principal axis  
• A ray which travel exactly along the principal axis will also satisfy the two conditions. So it is also included in the diagram
(b) Concave lens
• The definition is:
Light rays parallel and close to the principal axis diverge from one another after refraction. These rays appear to originate from a point on the same side. This point is the principal focus of a concave lens. It is denoted by the letter 'F'
• This will be clear from fig.7.13(b) above.
5. Focal length

It is the distance between the principal focus F and the optic centre P. This distance is denoted by the letter 'f'

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■ Why do we say that the principal focus of a concave lens is virtual?
Let us analyse:
• Consider the dashed green lines diverging from F in the fig.7.13(b). 
• The lines are dashed from 'F' up to the outer boundary of the lens. 
• Why are they 'dashed'?
• The answer can be written in steps: 
1. The actual light rays pass through the lens and after refraction, leaves the outer boundary of the lens
• They diverge from each other
2. These rays cannot travel back to give us the position of 'F'
• That is., there are no actual rays along the dashed lines behind an actual concave lens in 'real life'
    ♦ So they are drawn with dashed lines
• Consequently, there is no actual 'F' behind an actual concave lens in 'real life' 
3. We can only draw them on paper.
• So the principal focus of a concave lens is said to be virtual
■ In short, we can write:
It is impossible to converge light at a point using a concave lens. Therefore the principal focus of a concave lens is virtual

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Now we will learn about image formation. Consider fig.7.14 below:

Fig.7.14

1. A convex lens is placed in front of a candle. A screen is placed in front of the lens. 
• In this arrangement we have two distances:
(i) Distance between the object (candle) and the lens. This is denoted by the letter 'u'
(ii) Distance between the image (screen) and the lens. This is denoted by the letter 'v'
2. By adjusting the above two distances, we can form a clear image of the candle on the screen

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■ How is the image formed on the screen?
We will write the answer in steps:
1. Consider the fig.7.15 below:

Fig.7.15

• An object (consisting of an yellow pole and a cube at the top) is placed on the left side of a lens
2. We know that rays of light reflect from the object in all directions. When those rays fall on our eyes, we see the object. 
• Two such rays are considered here. Those two rays start from the cube and travel in two different directions. 
• That is., after starting from the cube, they diverge from each other. 
• But after passing through the lens, they converge at a point on the other side of the lens. 
3. At that point of convergence, we get an image of the cube.
■ So we can note a point:
To define a single point on the image, we will need at least two rays.
4. In the fig.7.15 above, we defined the cube in the image. In the same way, we can define all other points on the image.

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• 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 first familiarise ourselves with some rules. We will see those rules in the next section.

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