Question

Draw a ray diagram showing the formation of the image by a point object on the principal axis of a spherical convex surface separating two media of refractive indices $n_1$ and $n_2$, when a point source is kept in rarer medium of refractive index $n_1$. Derive the relation between object and image distance in terms of refractive index of the medium and radius of curvature of the surface.
Hence obtain the expression for lens-maker's formula in the case of thin convex lens.

Answer 1

Consider a convex lens (or concave lens) of absolute refractive index ${\mu }_2$ to be placed in a rarer medium of absolute refractive index ${\mu }_1$.

Considering the refraction of a point object on the surface $X{P}_{1}Y$, the image is formed at $I_1$ who is at a distance of $V_1$.

$C{I}_1=P_1{I}_1=V_1$ (as the lens is thin)

$C{C}_1=P_1{C}_1=R_1$

$CO=P_{1}O=u$

It follows from the refraction due to convex spherical surface $X{P}_{1}Y$

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

The refracted ray from A suffers a second refraction on the surface $X{P}_{2}Y$ and emerges along BI. Therefore I is the final real image of O.

Here the object distance is

$u=C{I}_1\approx P_2{I}_1=V$

$P_1{P}_2$ is very small

Let $CI\approx P_{2}I=V$

(Final image distance)

Let $R_2$ 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

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

Adding $\left(i\right)and\left(ii\right)$

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

or

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

But $\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$ and $\frac{{\mu }_2}{{\mu }_1}=\mu$

Thus we get=

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