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

• A. $-20cm$
• B. $-25cm$
• C. $-50cm$
• D. $50cm$

Answer 1

Answer: (C)

Given ${\mu }_g=1.5$

${\mu }_{\text{oil }}=1.7$
$R=20cm$
From Lens Maker's formula for the plano convex lens
$\frac{1}{f}=(\mu -1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right]$
Here, $R_1=R$
and for plane surface, $R_2=\infty$
$\therefore \frac{1}{f_{\text{lens }}}=(1.5-1)\left(\frac{1}{R}-0\right)$
$\Rightarrow \frac{1}{f_{\text{lens }}}=\frac{0.5}{R}$
When the intervening medium is filled with oil,
then focal length of the concave lens formed by the oil
$\frac{1}{f_{\text{concave }}}=(1.7-1)\left(-\frac{1}{R}-\frac{1}{R}\right)$
$=-0.7\times \frac{2}{R}=\frac{-1.4}{R}$
Here, we have two concave surfaces
So, $\frac{1}{f_{eq}}=2\times \frac{1}{f}+\frac{1}{f}=2\times \frac{0.5}{R}+\left(\frac{-1.4}{R}\right)$
$=\frac{1}{R}-\frac{1.4}{R}=-\frac{0.4}{R}$
$\therefore f_{eq}=-\frac{R}{0.4}=-\frac{20}{0.4}=-50cm$