Question

One plano-convex and one plano-concave lens of same radius of curvature $R$ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of $1$ is ${\mu }_1$ and that of $2$ is ${\mu }_2$ , then the focal length of the combination is

• A. $\frac{2R}{{\mu }_1-{\mu }_2}$
• B. $\frac{R}{2-({\mu }_1-{\mu }_2)}$
• C. $\frac{R}{2({\mu }_1-{\mu }_2)}$
• D. $\frac{R}{{\mu }_1-{\mu }_2}$

Answer 1

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

If two or more lenses are in contact, the equivalent focal length is given by the formula
$$\frac{1}{f}=\sum _{i=1}^n\frac{1}{f_i}$$
Here,
$$\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}$$
Focal length of an plano - convex lens is given by
$$f_1=\frac{R_1}{{\mu }_1-1}$$
Focal length of an plano - concave lens is given by
$$f_2=-\frac{R_2}{{\mu }_2-1}$$
From the figure and from the data given in the question we can deduce that,

$\frac{1}{f}=\left(\frac{{\mu }_1-1}{R}\right)-\left(\frac{{\mu }_2-1}{R}\right)$

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

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

Hence, option (d) is the correct answer.

Why this question? Note:- The given lens combination may converge or diverge light depending upon ${\mu }_1$ and ${\mu }_2$. Let's try:- Identify the behaviour of the given lens combination when, (a) ${\mu }_1>{\mu }_2$ , (b) ${\mu }_1<{\mu }_2$ and (c) ${\mu }_1={\mu }_2$