Question

Four combinations of two thin lenses are given in List - I. The radius of curvature of all curved surfaces is $r$ and the refractive index of the medium of all the lenses is $1.5$. Match the lens combination in List - I with their focal length in List - II and select the correct answer using the code given below the lists.

• A. $P-1,Q-2,R-3,S-4$
• B. $P-2,Q-4,R-3,S-1$
• C. $P-4,Q-1,R-2,S-3$
• D. $P-2,Q-1,R-3,S-4$

Answer 1

The correct option is B $P-2,Q-4,R-3,S-1$

Given that,
Radius of curvature of each lens, $R=r$
Refractive index of each lens, $\mu =1.5$

Focal length of a lens is given by
$\frac{1}{f}=(\frac{{\mu }_2}{{\mu }_1}-1)(\frac{1}{R_1}-\frac{1}{R_2})$

Focal length of single biconvex lens
$\frac{1}{f}=(\mu -1)[\frac{1}{R_1}-\frac{1}{R_2}]$
$\Rightarrow \frac{1}{f}=(1.5-1)[\frac{1}{r}-\frac{1}{-r}]$
$\Rightarrow \frac{1}{f}=\left(0.5\right)\left[\frac{2}{r}\right]$
$\Rightarrow \frac{1}{f}=\frac{1}{r}$ ...(i)
$\Rightarrow f=r$

Similarly,
Focal length of plano-convex lens $=2r$
& Focal length of plano-concave lens $=-2r$

For the combination $P$:
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{r}+\frac{1}{r}=\frac{2}{r}$
$\therefore F=\frac{r}{2}$

For combination $\text{Q}$: Combination of two plano-convex lenses.
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{2r}+\frac{1}{2r}=\frac{2}{2r}=\frac{1}{r}$
$\therefore F=r$

For combination $R$: Combination of two plano-concave lenses.
$\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=-\frac{1}{2r}-\frac{1}{2r}=-\frac{2}{2r}=-\frac{1}{r}$
$\therefore F=-r$

For combination $S$: One is a convex lens and another is a plano-concave lens.
$\frac{1}{F}=\frac{1}{r}+\frac{1}{-2r}$
$\Rightarrow \frac{1}{F}=\frac{1}{2r}$
$\Rightarrow F=2r$

Hence, option (b) is correct.