JEE PYQ: Ray Optics Question 46
Question 46 - 2019 (10 Jan Shift 1)
A plano convex lens of refractive index $\mu_1$ and focal length $f_1$ is kept in contact with another plano concave lens of refractive index $\mu_2$ and focal length $f_2$. If the radius of curvature of their spherical faces is $R$ each and $f_1 = 2f_2$, then $\mu_1$ and $\mu_2$ are related as:
(1) $\mu_1 + \mu_2 = 3$
(2) $2\mu_1 - \mu_2 = 1$
(3) $3\mu_2 - 2\mu_1 = 1$
(4) $2\mu_2 - \mu_1 = 1$
Show Answer
Answer: (2)
Solution
From lens maker’s formula, $\frac{1}{f} = (\mu - 1)\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$ $\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = \frac{1}{2f_2}$ Similarly, for plano-concave lens $\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right)$ Dividing $\frac{1}{f_1}$ by $\frac{1}{f_2}$ we get, $\frac{(\mu_1 - 1)}{R} = \frac{(\mu_2 - 1)}{2R}$ or, $2\mu_1 - \mu_2 = 1$
BETA
