Question

What is a lens maker's fromula? Why it is called so? Derive it for a thin convex lens.

Or

Show by a diagram the image formation of a point object by a thin double convex lens having radii of  curvature R₁ and R₂. Hence derive the formula \(\frac{1}{f} = (n-1)[\frac{1}{R_1} -\frac{1}{R_2}],\) where f is the focal length and n is refractive index of material of the lens.

Answer 1

 Lens maker’s formula

It is given by

\(\frac{1}{f} = (\mu - 1)(\frac{1}{R_1} -\frac{1}{R_2})\)

This formula relates the focal length of the lens of the radii of curvature of the surfaces and the refractive index of the material of the lens.

Why is it called lens maker’s formula?

It is so called because this formula is used by the manufacturers to prepare lenses of desired focal lengths.

Sign conventions

• All the distances are measured from the optical centre of lens.
• Distance measured in the direction of incident light is taken +ve.
• Distance measured in the direction opposite to the direction of incident light is taken -ve.

Assumptions

1. Object is a point object.
2. Lens is thin so that distances are measured from the poles of its surfaces.
3. The aperture of the lens is small.
4. Angles of incidence and refraction are small.

Let C₁ and C₂ be centres of curvature of two spherical surfaces XP₁Y and XP₂Y with optical centre C. Let µ₂ be the refractive index of the material of the lens and µ₁ that of the medium surrounding the lens where µ₂ > µ₁.

Refraction at surface XP₁Y

Let a ray OA starts from the point object O which is placed on the principal axis of the convex lens and it refracts into the denser medium at point A. The refracted ray AB goes undeviated to meet at point I₁, if the lens material were continued. Then I₁ would be the real image of point object O.
From Snell’s law for small i₁ and r₁, we have

\(\frac{sin\ i}{\sin r_1} = \frac{i_1}{r_1} = \frac{\mu_2}{\mu_1}\)

or µ₁ i₁ = µ₂ r₁ ..........(1)

In ∆ AOC1,

i₁ = α₁ + γ₁

In ∆ AI₁C₁,

γ₁ = ß₁ + r₁

or r₁ = γ₁ - ß₁

Putting the values of i₁ and r₁ in Eq.(1), we get

µ₁ (α₁ + γ₁) = µ₂ (γ₁ - ß₁)

or µ₁α₁ + µ₁γ₁ = µ₂γ₁ - µ₂ß₁

or µ₁α₁ + µ₂ß¹ = (µ₂ - µ₁) γ¹

Since α₁, ß₁ and γ₁ are small, so they can be replaced by their tangents.

∴ µ₁ tan α₁ + µ₂ tan ß₁ = (µ₂ - µ₁) tan γ₁

Refraction at surface XP₂Y

Actually the lens is not continuous and the ray AB suffer refraction at B and finally meet at I. So I is the real image of the virtual object I₁ (placed in the material of the lens).
From Snell’s law for small i₂ and r₂, we have

\(\frac{sin\ i_2}{\sin r_2} = \frac{i_2}{r_2} = \frac{\mu_1}{\mu_2}\)

or µ₁r₂ = µ₁i₂ ..........(3)

In ∆ BC₂I₁,

i₂ = ß₁ + γ₂

In ∆ BC₂I, r² = ß² + γ²

Putting the value of i₂ and r₂ in Eq. (3), we get

µ₁ (ß₂ + γ₂) = µ₂ (ß₁ + γ₂)

or µ₁ß₂ + µ₁γ₂ = µ₂ß₁ + µ₂γ₂

or µ₁ß₂ - µ2ß₁ = (µ₂ - µ₁) γ₂

Since ß₂, ß₁ and γ₂ are small, so they can be replaced by their tangents, so

µ₁ tan ß₂ - µ₂ tan ß₁ = (µ₂ - µ₁) tan γ₂