Question

A plano convex lens of refractive index${\mu }_1$ and focal length $f_1$ is kept in contact with another plano concave lens of refractive index ${\mu }_2$ and focal length $f_2$. If the radius of curvature of their spherical faces is $R$ each and $f_1=2f_2$, then ${\mu }_1$ and ${\mu }_2$ are related as :

• A. ${\mu }_1+{\mu }_2=3$
• B. $2{\mu }_2-{\mu }_1=1$
• C. $2{\mu }_1-{\mu }_2=1$
• D. $3{\mu }_2-2{\mu }_1=1$

Answer 1

The correct option is C

$2{\mu }_1-{\mu }_2=1$

Step 1: Given data

Refractive index of plano convex lens$={\mu }_1$

Focal length of convex surface of plano convex lens $=f_1$

Refractive index of plano concave lens$={\mu }_2$

Focal length of concave surface of plano concave lens$=f_2$

The radius of curvature of their spherical faces$=R$

Relation between $f_1$ and $f_2$ is given by following relation

$f_1=2f_2$

Step 2. Find the relation between ${\mathit{\mu }}_{\mathbf{1}}$ and ${\mathit{\mu }}_{\mathbf{2}}$

Using lens maker formula

$\frac{1}{f_{}}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$

For plano convex lens

$$\begin{gathered} \frac{1}{f_1}=({\mu }_1-1)[\frac{1}{\infty }-(\frac{1}{-R}\left)\right] \\ \frac{1}{f_1}=({\mu }_1-1)\left(\frac{1}{R}\right) \\ {f}_1=\frac{R}{({\mu }_1-1)} \end{gathered}$$

For plano concave lens

$$\begin{gathered} \frac{1}{f_2}=({\mu }_2-1)\left[\right(\frac{1}{-R})-\frac{1}{\infty }] \\ \frac{1}{f_2}=({\mu }_2-1)\left(\frac{1}{-R}\right) \\ {f}_2=\frac{-R}{({\mu }_2-1)} \end{gathered}$$

As,

$$\begin{gathered} 2\left|f_2\right|=\left|f_1\right| \\ \frac{2R}{({\mu }_2-1)}=\frac{R}{({\mu }_1-1)} \\ 2R({\mu }_1-1)=R({\mu }_2-1) \\ 2{\mu }_1-2={\mu }_2-1 \\ 2{\mu }_1-{\mu }_2=1 \end{gathered}$$

Hence, option (C) is correct.