Question

A plano convex lens (focal length $f_2$, 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. Then, the focal length of the combination will be

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

Answer 1

The correct option is B

$\frac{R}{{\mu }_2-{\mu }_1}$

Given combination is as shown below





As lenses are in contact, equivalent focal length of combination is
$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}$
Using lens Maker's formula,

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

Here, $\frac{1}{f_1}=({\mu }_1-1)(\frac{1}{-R}-\frac{1}{\infty })=\frac{(1-{\mu }_1)}{R}$

and $\frac{1}{f_2}=({\mu }_2-1)(\frac{1}{\infty }-\frac{1}{(-R)})=\frac{({\mu }_2-1)}{R}$

$\therefore$ $\frac{1}{f_{eq}}=\left(\frac{{\mu }_2-1}{R}\right)+\left(\frac{1-{\mu }_1}{R}\right)$

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

So, $f_{eq}=\frac{R}{{\mu }_2-{\mu }_1}$

Why this question? Importance in JEE: The questions based on Lens Maker's formula, $\frac{1}{f}=(\mu -1)(\frac{1}{R_1}-\frac{1}{R_2})$ are common in JEE and other competitive exams.