Lens Maker’s Formula: Derivation and Solved Examples

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What is the Derivation of Lens Makers Formula for Concave Lens?

Newton’s Equation

Let LL’ be a convex lens. F, F’ and O are the second principal focus, the first principal focus and the optical centre of the lens respectively [Fig.], P and Q are the point object and point image respectively.

Here, OF = OF’ = f; PF’ = x and QF = y; object distance, OP = -u = -(x + f) and OQ = v = y + f.
The equation of the lens is
\(\frac{1}{v}-\frac{1}{u}\) = \(\frac{1}{f}\)
∴ \(\frac{1}{y+f}+\frac{1}{x+f}\) = \(\frac{1}{f}\) or, \(\frac{x+f+y+f}{(y+f)(x+f)}\) = \(\frac{1}{f}\)
or (y + f)(x + f) = f(x + y + 2f)
or xy + xf + yf + f² = xf + yf + 2f² or, xy = f²
This is Newton’s equation in case of lens.
For any lens, f is a constant quantity. Hence, a x – y graph supporting Newton’s equation will be a rectangular hyperbola.

Lens Maker’S Formula For Thin Lens

The lens maker’s formula involves the focal length, refractive index of the material of the lens and the radius of curvature of the two surfaces of the lens. This formula is derived from the refraction of light on the two spherical surfaces of a lens.

In Fig. refraction of light at the two spherical surfaces of a biconvex lens has been shown. P is the point object on the principal axis of the lens; Q’ is the image formed due to refraction of light at the first surface of the lens and Q is the final image.
Let µ₁ = refractive index of the medium in which the object is placed
µ₂ = refractive index of the material of the lens
µ₃ = refractive index of the medium into which the final ray emerges

In Gaussian system, object distance is measured from the pole O of the first spherical surface and final image distance is measured from the pole O’ of the second spherical surface. Accordingly,
object distance, OP = -u
image distance, OQ’ = v’
final image distance, O’Q = v
thickness of lens on the axis, 00′ = t
radius of curvature of the first surface of the lens = r₁
radius of curvature of the second surface of the lens = -r₂
Considering refraction at the first surface AOB of the lens we have,
\(\frac{\mu_2}{v^{\prime}}-\frac{\mu_1}{u}\) = \(\frac{\mu_2-\mu_1}{r_1}\) …. (1)
[both object and image are real]
Considering refraction at the second surface AO’B of the lens we have,
\(\frac{\mu_3}{v}-\frac{\mu_2}{v^{\prime}-t}\) = \(\frac{\mu_3-\mu_2}{r_2}\) …. (2)
[object is virtual and image is real]
If the lens is very thin i.e., t\(\ll\)v’, then t can be neglected.
In that case equation (2) becomes
\(\frac{\mu_3}{v}-\frac{\mu_2}{v^{\prime}}\) = \(\frac{\mu_3-\mu_2}{r_2}\) ….. (3)
Adding equations (1) and (3) we have,
\(\frac{\mu_3}{v}-\frac{\mu_1}{u}\) = \(\frac{\mu_2-\mu_1}{r_1}\) + \(\frac{u_3-\mu_2}{r_2}\) …. (4)

This is the general equation of lens. This formula has been obtained for formation of real image by a convex lens. But this formula is equally applicable for the formation of virtual image by a convex lens or for the concave lens.
If the surrounding medium be air, then µ₁ = µ₃ = 1. Taking µ₂ = µ for the r.i. of the material, we have
\(\frac{1}{v}-\frac{1}{u}\) = (µ – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\) …. (5)
If the object is at infinity, image will be formed at the principal focus.
i.e., if u = ∞, v = f
∴ \(\frac{1}{f}\) = (µ – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\) …. (6)
This is the lens maker’s formula.

Lens in air: If the refractive index of a glass lens relative to air is _aµ_g and the radii of curvature of the first and the second refracting surfaces are r₁ and r₂ respectively, focal length f of the lens is obtained from the following relation,
\(\frac{1}{f}\) = _aµ_g\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\) …. (7)

i) In case of biconvex lens : r₁ is positive and r₂, is negative, So for this lens from equation (7) we have,
\(\frac{1}{f}\) = (_aµ_g – 1)\(\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\) …. (8)
If the lens is equi-convex, then r₁ = r₂ = r and in that case,
\(\frac{1}{f}\) = (_aµ_g – 1)\(\frac{2}{r}\) ….. (9)

ii) In case of biconcave lens: r₁ is negative and r₂ is positive. So for this lens from equation (7) we have,
\(\frac{1}{f}\) = – (_aµ_g – 1)\(\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\) …. (10)
If the lens is equi-concave, then r₁ = r₂ = r. In this case,
\(\frac{1}{f}\) = -(_aµ_g – 1) ᐧ \(\frac{2}{r}\) ……. (11)

Dependence of focal length on surrounding medium: Let us suppose that a lens is situated in a medium denser than air. Suppose, the denser medium is water.

Now, if the focal lengths of the lens in air and water are f_a and f_w respectively, then
\(\frac{1}{f_a}\) = (_wµ_g – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\) and \(\frac{1}{f_w}\) = (_wµ_g – 1)\(\left(\frac{1}{r_1}-\frac{1}{r_2}\right)\)
[r₁ and r₂ are the radii of curvature of the first and the
second refracting surfaces respectively]

So, focal length of a lens increases with the increase of optical density of the surrounding medium.

If a convex lens of focal length f is cut horizontally along its principal axis into two equal halves, each half will have focal length equal to f because the radii of curvature of the two sur-faces of the new parts have the same values as the original lens.

Numerical Examples

Example 1.
The focal length of a glass-lens in air is 5 cm. What will be its focal length in water? Refractive index of glass = 1.51 and refractive index of water = 1.33.
Solution:
Let focal length of the lens in air = f_a ; radii of curvature of the two surfaces = r₁ and r₂

Example 2.
A convex lens (µ = 1.5) is immersed in water (µ = 1.33). Will the focal length of the lens change in water? If so, how?
Solution:
If the focal length of the convex lens in air is f_a, refractive index of the material of the lens is _aµ_g and radii of curvature of the two surfaces are r₁ and r₂, then following sign convention
\(\frac{1}{f_a}\) = +(_aµ_g – 1)\(\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\) …. (1)
If the lens is immersed in water, focal length of the lens will be changed. If the focal length of the lens when immersed in water is f_w, then
\(\frac{1}{f_w}\) = +(_aµ_g – 1)\(\left(\frac{1}{r_1}+\frac{1}{r_2}\right)\) …. (2)
Dividing equation (1) by equation (2)

So, if the lens is immersed in water its focal length will be about 4 times its focal length in air.

Example 3.
The radii of curvature of two surfaces of a biconvex glass-lens are 20 cm and 30 cm. What is its focal length in air and water? Refractive index of glass = \(\frac{3}{2}\), refractive index of water = \(\frac{4}{3}\).
Solution:
If the focal length of the lens in air is f_a and following sign convention, we have

Therefore, the focal length of the biconvex lens in air is 24 cm . If the focal length of the lens in water is f_w, we have

So, the focal length of the lens in water is 96 cm .

Example 4.
A plano-convex lens has radius of curvature 10 cm. It focal length is 80 cm in water. Calculate the refractive index of the material of the lens. Given refractive index
of water = \(\frac{4}{3}\). [WBCHSE Sample Question]
Solution:
Let absolute refractive index of the material of the lens = n ; r.i. of the material with respect to water = n₁; absolute r.i. of water = n’.
So, n₁ = \(\frac{n}{n^{\prime}}\) or, n = n₁n’.
Given, r₁ = ∞ and r₂ = -10 cm. The focal length of the plano-convex lens when immersed in water = 80 cm.

Example 5.
The refractive index of the material of an equi-convex lens is 1.5 and radius of curvature of each spherical surface is 20 cm. Calculate the focal length of the lens.
[HS’13]
Solution:
Let the focal length of the lens be f and radius of curvature of each spherical surface be r.
∴ \(\frac{1}{f}\) = (µ – 1) ᐧ \(\frac{2}{r}\) = (1.5 – 1) × \(\frac{2}{20}\) = \(\frac{1}{20}\)
∴ f = 20 cm