Q. Shown in the figure here is a convergent lens placed inside a cell filled with a liquid. The lens has focal length + 20 $cm$ when in air and its material has refractive index 1.50. If the liquid has refractive index 1.60, the focal length of the system is
A) $+ 80\ cm$
B) $- 80\ cm$
C) $- 24\ cm$
D) $- 100\ cm$
Explanation
(d) $\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} + \frac{1}{f_{3}}$
$\frac{1}{f_{1}} = (1.6 - 1)\left( \frac{1}{\infty} - \frac{1}{20} \right) = - \frac{0.6}{20} = - \frac{3}{100}\ \ \ldots(i)$
$\frac{1}{f_{2}} = (1.5 - 1)\left( \frac{1}{20} - \frac{1}{- 20} \right) = \frac{1}{20}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ldots(ii)$
$\frac{1}{f_{3}} = (1.6 - 1)\left( \frac{1}{- 20} - \frac{1}{\infty} \right) = - \frac{3}{100}\ \ \ \ \ \ \ \ \ \ \ \ \ldots(iii)$
$\Rightarrow \frac{1}{F} = - \frac{3}{100} + \frac{1}{20} - \frac{3}{100} \Rightarrow F = - 100\ cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.