Question

Write the formula for the refraction of light on a spherical surface. Establish lens formula $\frac{1}{f}=(n-1)(\frac{1}{R_1}-\frac{1}{R_2})$ with its help. Symbols used their usual meanings.

Answer 1

The following assumptions are made for the derivation:

1) The lens is thin, so that distances measured from the poles of its surfaces can be taken as equal to the distances from the optical centre of the lens.

2) The aperture of the lens is small.

3) Point object is considered.

4} Incident and refracted rays make small angles

Consider a convex lens (or concave lens) of absolute refractive index m2 to be placed in a rarer medium of absolute refractive index m1.

Considering the refraction of a point object on the surface XP1Y, the image is formed at I1 who is at a distance of V1.

CI1= P1I1 = V1 (as the lens is thin)

CC1 = P1C1 = R1

CO = P1O = u

It follows from the refraction due to convex spherical surface XP1Y

$\frac{{\mu }_1}{-u}+\frac{{\mu }_2}{v_1}=\frac{{\mu }_2-{\mu }_2}{R_1}....\left(1\right)$

The refracted ray from A suffers a second refraction on the surface XP2Y and emerges along BI. Therefore I is the final real image of O.

Here the object distance is

(Note P1P2 is very small)

(Final image distance)

Let R2 be radius of curvature of second surface of the lens.

It follows from refraction due to concave spherical surface from denser to rarer medium that:-

$\frac{-{\mu }_2}{v_1}+\frac{{\mu }_1}{v}=\frac{{\mu }_1-{\mu }_2}{R_2}....\left(2\right)$

Adding (1) & (2) we get:-

$\frac{{\mu }_1}{-u}+\frac{{\mu }_1}{v}=({\mu }_2-{\mu }_1)(\frac{1}{R_1}-\frac{1}{R_2})$

$\frac{1}{v}-\frac{1}{u}=\frac{({\mu }_2-{\mu }_1)}{{\mu }_1}\times (\frac{1}{R_1}-\frac{1}{R_2})$

and we know,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$ and $\frac{{\mu }_2}{{\mu }_1}{=}_1{\mu }_2=\mu \left(say\right)$

therefore,

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