Question

A plano-convex lens (focal length ${\int }_2$ refractive index ${\mu }_2$ radius of curvature R) fits exactly into a plano-concave lens (focal length $f_1$ refractive index ${\mu }_1$ radius of curvature R). Their plane surfaces are parallel to each other. the focal of the will be :

• A. $f_1-f_2$
• B. $\frac{R}{{\mu }_2-{\mu }_1}$
• C. $\frac{2f_1{f}_2}{f_1+f_2}$
• D. $f_1+{\int }_2$

Answer 1

The correct option is B $\frac{R}{{\mu }_2-{\mu }_1}$

Focal length of the combination

$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$ ………….$\left(1\right)$

We have,

$f_1=\frac{R}{({\mu }_1-1)}$ $f_2=\frac{R}{({\mu }_2-1)}$

$\Rightarrow \frac{1}{f_1}=\frac{{\mu }_1-1}{R}$ $\Rightarrow \frac{1}{f_2}=\frac{{\mu }_2-1}{R}$

Putting these values in equation $\left(1\right)$

$\frac{1}{f}=\frac{{\mu }_1-}{R}-\frac{{\mu }_2-1}{R}$

$\frac{1}{f}=\frac{{\mu }_1-1-{\mu }_2+1}{R}=\frac{{\mu }_1+{\mu }_2}{R}$

$\Rightarrow f=\frac{R}{{\mu }_1-{\mu }_2}$.