Question

Two identical thin plano-convex glass lenses (refractive index $1.5$) each having radius of curvature of $20cm$ are placed with their convex surfaces in contact at the centre. The intervening space is filled with oil of refractive index $1.7$. The focal length of the combination is

Answer 1

Here the light passes through three refracting media, $\text{I, II \& III}$ as shown in the figure.

Media $\text{I \& II}$ has a refractive index $1.5$ and radius of curvature $20cm$. While medium $\text{II}$ has a refractive index of $1.7$ and it acts as a concave lens.

Let the focal length of the three media be $f_1,f_2,f_3$.

Applying the lens makers formula

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

For media $\text{I}$,

$R_1=\infty$ since it is a plano convex lens.

$\frac{1}{f_1}=\left(1.5-1\right)\left(\frac{1}{\infty }-\frac{1}{-20}\right)=0.5\times \frac{1}{20}=\frac{1}{40}$

For media $\text{II}$,

$\frac{1}{f_2}=\left(1.7-1\right)\left(\frac{1}{-20}-\frac{1}{20}\right)=0.7\times \left(-\frac{2}{20}\right)=-\frac{0.7}{10}$

For media $\text{III}$,

Since media $\text{I \& III}$ are similar we can write

$\frac{1}{f_3}=\frac{1}{40}$

For applying the formula for equivalents lenses,

$\frac{1}{f_{eq}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}$

$\Rightarrow \frac{1}{f_{eq}}=\frac{1}{40}-\frac{0.7}{10}+\frac{1}{40}$

$\Rightarrow \frac{1}{f_{eq}}=\frac{1-2.8+1}{40}=-\frac{0.8}{40}=-\frac{1}{50}$

$\therefore f_{eq}=-50cm$

Hence the correct answer is option (D).