Question

Two identical thin plano-convex glass lenses (refractive index $1.5$) each having radius of curvature of $20 \,cm$ 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. $-20 \,cm$
• B. $-25 \, cm$
• C. $-50 \, cm$
• D. $50 \,cm$

Answer 1

Answer: (C)

Given $\mu_{g}=1.5$

$\mu_{\text {oil }} =1.7 $
$ R =20 \,cm $
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, $\frac{1}{f_{ eq }}=2 \times \frac{1}{f}+\frac{1}{f}=2 \times \frac{0.5}{R}+\left(\frac{-1.4}{R}\right)$
$=\frac{1}{R}-\frac{1.4}{R}=-\frac{0.4}{R}$
$\therefore f_{ eq }=-\frac{R}{0.4}=-\frac{20}{0.4}=-50\, cm$