Question

A thin lens made of a material of refractive index ${\mu }_2$, has a medium of refractive index ${\mu }_1$ on one side and a medium of refractive index ${\mu }_2$ on the other side. The lens is biconvex and the two radii of curvature have equal magnitude R. A beam of light travelling parallel to the principal axis is incident on the lens. Where will the image be formed if the beam is incident from

(a) The medium ${\mu }_1$, and

(b) From the medium ${\mu }_3$ ?

Answer 1

(a) For 1st surface,

$\frac{{\mu }_2}{V}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}\dots \dots \left(A\right)$

Image for surface (i) will be object for surface (ii).

$\frac{{\mu }_3}{V_{\text{final}}}-\frac{{\mu }_2}{V}=\frac{{\mu }_3-{\mu }_2}{R}\dots \dots \left(B\right)$

Equation (A) + (B)

$\frac{{\mu }_3}{V_{\text{final}}}-\frac{{\mu }_1}{u}=\frac{{\mu }_2-{\mu }_1}{R}-\frac{{\mu }_3-{\mu }_2}{R}$

In case a u = a

So $\frac{{\mu }_3}{V_{\text{final}}}=\frac{2{\mu }_2-{\mu }_1-{\mu }_3}{R}$

$V_{\text{final}}=\frac{{\mu }_{3}R}{2{\mu }_2-{\mu }_1-{\mu }_3}$

(b)

For 1st surface,

$\frac{{\mu }_2}{V}-\frac{{\mu }_3}{u}=\frac{{\mu }_2-{\mu }_3}{R}$

(i) will be object for surface (ii)

For IInd surface,

$\frac{{\mu }_1}{V_{\text{final}}}-\frac{{\mu }_2}{V}=\frac{{\mu }_1-{\mu }_2}{-R}$

Equation (A) + (B)

$\frac{{\mu }_1}{V_{\text{final}}}-\frac{{\mu }_2-{\mu }_3}{R}=\frac{({\mu }_1-{\mu }_2)}{R}$

$V_{\text{final}}=\frac{{\mu }_{1}R}{2{\mu }_2-{\mu }_1-{\mu }_3}$