Solution

$R_1 = +20$ cm, $R_2 = -20$ cm, $\mu = \frac{3}{2}$ $P = \frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $P = \left(\frac{3}{2} - 1\right)\left(\frac{1}{0.2} + \frac{1}{0.2}\right) = \frac{1}{2}\left(\frac{2}{0.2}\right) = +5$ D

Concept power of a biconvex lens

Ray Optics and Optical Instruments in past papers

89 questions from this chapter have appeared across 25 exam years.