Question

A lens of particular focal length is made from a given glass by adjusting radius of curvature. The formula applied in this case is lens maker’s formula

1. Write down lens maker’s formula

2. Derive lens maker’s formula considering refraction at a spherical surface

3. Explain the following facts based on lens maker’s formula

• power of sun glasses is zero even though they are curved
• if a lens is immersed in water focal length increases

Answer 1

1. \(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

2. Refraction by a lens:

Lens Maker’s Formula (for a thin lens): Consider a thin lens of refractive index n₂ formed by the spherical surfaces ABC and ADC. Let the lens is kept in a medium of refractive index n₁ . Let an object ‘O’ is placed in the medium of refractive index n₁ . Hence the incident ray OM is in the medium of refractive index n₁ and the refracted ray MN is in the medium of refractive index n₂ .

The spherical surface ABC (radius of curvature R,) forms the image at l₁ . Let ‘u’ be the object distance and ‘v₁ ’ be the image distance.

Then we can write,

\(\frac{n_2}{v_1}-\frac{n_1}{u}=\frac{n_2-n_1}{R_1}\) ......(1)

Adding eq (1) and eq (2) we get

\(\frac{n_2}{v_1}-\frac{n_1}{u}+\frac{n_1}{v}-\frac{n_2}{v_1}\) = \(\frac{n_2-n_1}{R_1}+\frac{n_1-n_2}{R_2}\)

\(\frac{n_1}{v}-\frac{n_1}{u}=(n_2-n_1)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) 

 Dividing throughout by n₁ , we get

\(\frac{1}{v}-\frac{1}{u}=\Big(\frac{n_2}{n_1}-1\Big)\)\(\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

if the lens is kept in air ,\(\frac{n_2}{n_1}\) = n

So the above equation can be written as,

\(\frac{1}{v}-\frac{1}{u}\)\((n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(4)

From the definition of the lens, we can take, when u = ∞, f = v

Substituting these values in the eq (3), we get

∴ \(\frac{1}{f}-\frac{1}{∞}\)\(=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

This is lens maker’s formula

\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\) ......(5)

For convex lens.

f = +ve, R = +ve, R = – ve

\(\frac{1}{f}=(n-1)\Big(\frac{1}{R_1}+\frac{1}{R_2}\Big)\)

For concave lens,

f = -ve, R = -ve, R = +ve

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

Lens formula

From eq(4),

\(\frac{1}{v}-\frac{1}{u}=(n-1)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

From eq(5)

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

From these two equations, we get

\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)

Linear magnification :

If h is the height of the object and h_o is the height of the image, then linear magnification.

\(m=\frac{h_1}{h_o}=\frac{v}{u}\)

3. a. R₁ = R, R₂ = +R

∴ \(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R}-\frac{1}{R}\Big)\)

power of lens, P = 0

b. We know

\(\frac{1}{f}=\Big(\frac{n_2}{n_1}-1\Big)\Big(\frac{1}{R_1}-\frac{1}{R_2}\Big)\)

The above equation shows when n₁ increases f decreases the refractive index of water is greater than air. Hence when we place a lens in water, focul length decreases.