Q. A thin equiconvex lens of refractive index 3/2 and radius of curvature 30 m is put in water (refractive index =$\frac{4}{3}).$ Its focal length is
A) 0.15 m
B) 0.30 m
C) 0.45 m
D) 1.20 m
Explanation
(d) $\frac{1}{f} = \left( \frac{\mu_{1}}{\mu_{2}} - 1 \right)\left( \frac{1}{R_{1}} - \frac{1}{R_{2}} \right)$
$\frac{1}{f} = \left( \frac{3/2}{4/3} - 1 \right)\left( \frac{1}{0.3} + \frac{1}{0.3} \right)$
Or $\frac{1}{f} = \left( \frac{9}{8} - 1 \right)\left( \frac{2}{0.3} \right)$
Or $\frac{1}{f} = \frac{1}{8} \times \frac{2}{0.3}\ \ $or $f = 1.20\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.