JEE PYQ: Ray Optics Question 39
Question 39 - 2019 (10 Apr Shift 1)
One plano-convex and one plano-concave lens of same radius of curvature ‘R’ but of different materials are joined side by side as shown in the figure. If the refractive index of the material of 1 is $\mu_1$ and that of 2 is $\mu_2$, then the focal length of the combination is:
(1) $\frac{R}{\mu_1 - \mu_2}$
(2) $\frac{2R}{\mu_1 - \mu_2}$
(3) $\frac{2R}{2(\mu_1 - \mu_2)}$
(4) $\frac{R}{2 - (\mu_1 - \mu_2)}$
Show Answer
Answer: (1)
Solution
Focal length of plano-convex lens: $\frac{1}{f_1} = (\mu_1 - 1)\left(\frac{1}{\infty} - \frac{1}{-R}\right) = \frac{\mu_1 - 1}{R}$ $\Rightarrow f_1 = \frac{R}{(\mu_1 - 1)}$ Focal length of plano-concave lens: $\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{1}{-R} - \frac{1}{\infty}\right) = \frac{\mu_2 - 1}{R}$ $\Rightarrow f_2 = \frac{-R}{(\mu_2 - 1)}$ For the combination of two lens: $\frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{\mu_1 - 1}{R} - \frac{\mu_2 - 1}{R}$ $= \frac{\mu_1 - \mu_2}{R}$ $\Rightarrow f_{\text{eq}} = \frac{R}{\mu_1 - \mu_2}$
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