JEE PYQ: Ray Optics Question 51
Question 51 - 2019 (12 Jan Shift 1)
A plano-convex lens (focal length $f_2$, refractive index $\mu_2$, radius of curvature R) fits exactly into a plano-concave lens (focal length $f_1$, refractive index $\mu_1$, radius of curvature R). Their plane surfaces are parallel to each other. Then, the focal length of the combination will be:
(1) $f_1 - f_2$
(2) $\frac{R}{\mu_2 - \mu_1}$
(3) $\frac{2f_1 f_2}{f_1 + f_2}$
(4) $f_1 + f_2$
Show Answer
Answer: (2)
Solution
$\frac{1}{f_2} = (\mu_2 - 1)\left(\frac{+1}{R}\right)$ $\frac{1}{f_1} = (\mu_1 - 1)\frac{(-1)}{R}$ Now when combined the focal length is given by $\frac{1}{f} = \frac{1}{f_1} + \frac{1}{f_2}$ $= (\mu_1 - 1)\frac{(-1)}{R} + (\mu_2 - 1)\frac{+1}{R}$ $= \frac{1}{R}[\mu_2 - f - \mu_1 + f]$ $= \frac{\mu_2 - \mu_1}{R}$ $\Rightarrow f = \frac{R}{\mu_2 - \mu_1}$
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