### Neet Solved Paper 2015 Question 36

##### Question: Two identical thin plano-convex glass lenses (refractive index 1.5) each having radius of curvature of 20 cm are placed wait 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

#### Options:

A) $ -20,cm $

B) $ -25,cm $

C) $ -50cm $

D) $ 50cm $

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

Given $ {\mu _g}=1.5 $ $ {\mu _{oil}},=1.7 $ $ R=20cm $

From Lens Maker’s formula for the piano convex lens $ \frac{1}{f}=(\mu -1)[ \frac{1}{R _1}-\frac{1}{R _2} ] $

Here, $ R _1=R $ and for plane surface $ R _2=\infty $

$ \therefore $ $ \frac{1}{f _{lens}}=(1.5-1)( \frac{1}{R}-0 ) $

$ \Rightarrow $ $ \frac{1}{f _{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 _{concave}}}=(17-1)-( -\frac{1}{R}-\frac{1}{R} ) $ $ =-0.7\times \frac{2}{R}=\frac{-14}{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}+( \frac{-14}{R} )=\frac{1}{R}-\frac{14}{R}=-\frac{0.4}{R} $

$ \therefore $ $ f _{eq}=-\frac{R}{0.4}=-\frac{20}{0.4}=-50cm $