Q. A thin lens made of glass of refractive index 1.5 has a front surface $+ 11\ D$ power and back surface $- 6\ D$. If this lens is submerged in a liquid of refractive index 1.6, the resulting power of the lens is
A) $- 0.5\ D$
B) $+ 0.5\ D$
C) $- 0.625\ D$
D) $+ 0.625\ D$
Explanation
(c) Total power $P = P_{1} + P_{2} = 11 - 6 = 5\ D$
Also $\frac{f_{l}}{f_{a}} = \frac{(\ _{a}\mu_{g} - 1)}{(\ _{l}\mu_{g} - 1)} \Rightarrow \frac{P_{a}}{P_{l}} = \frac{(\ _{a}\mu_{g} - 1)}{(\ _{l}\mu_{g} - 1)}$
$\Rightarrow \frac{5}{P_{l}} = \frac{(1.5 - 1)}{(1.5/1.6 - 1)} \Rightarrow P_{l} = - 0.625\ D$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.