JEE PYQ: Ray Optics Question 24
Question 24 - 2020 (06 Sep Shift 2)
A double convex lens has power $P$ and same radii of curvature $R$ of both the surfaces. The radius of curvature of a surface of a plano-convex lens made of the same material with power 1.5 $P$ is:
(1) $2R$
(2) $\frac{R}{2}$
(3) $\frac{3R}{2}$
(4) $\frac{R}{3}$
Show Answer
Answer: (4)
Solution
Given, using lens maker’s formula $\frac{1}{f} = (k - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ Here, $R_1 = R_2 = R$ (For double convex lens) $\therefore \frac{1}{f} = (\mu - 1)\left(\frac{1}{R} + \frac{1}{R}\right)$ $\Rightarrow P = \frac{1}{f} = (\mu - 1)\frac{2}{R}$ …(i) For plano convex lens, $R_1 = R’$, $R_2 = \infty$ Using lens maker’s formula again, we have $1.5P = (\mu - 1)\left(\frac{1}{R’} - \frac{1}{\infty}\right)$ …(ii) $\Rightarrow \frac{3}{2}P = \frac{\mu - 1}{R’}$ From (i) and (ii), $\frac{3}{2} = \frac{R’}{2R} \Rightarrow R’ = \frac{R}{3}$
BETA
