Q. The refractive index of a material of a planoconcave lens is 5/3, the radius of curvature is 0.3 m. The focal length of the lens in air is
A) -0.45m
B) -0.6m
C) -0.75m
D) -1.0m
Explanation
(a) Lens maker’s formula
$\frac{1}{f} = (\mu - 1)\ \left\lbrack \frac{1}{R_{1}} - \frac{1}{R_{2}} \right\rbrack$
Where , $R_{2} = \ \infty,\ R_{1} = 0.3\ m$
$\therefore\frac{1}{f} = \left( \frac{5}{3} - 1 \right)\left( \frac{1}{0.3} - \frac{1}{\infty} \right)$
$\Rightarrow \frac{1}{f} = \frac{2}{3} \times \frac{1}{0.3}$
$Or\ f = 0.45\ m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.