JEE PYQ: Ray Optics Question 38
Question 38 - 2019 (09 Apr Shift 2)
A thin convex lens L (refractive index $= 1.5$) is placed on a plane mirror M. When a pin is placed at A, such that OA $= 18$ cm, its real inverted image is formed at A itself, as shown in figure. When a liquid of refractive index $\mu_1$ is put between the lens and the mirror, the pin has to be moved to A’, such that OA’ $= 27$ cm, to get its inverted real image at A’ itself. The value of $\mu_1$ will be:
(1) $\frac{4}{3}$
(2) $\frac{3}{2}$
(3) $\sqrt{3}$
(4) $\sqrt{2}$
Show Answer
Answer: (1)
Solution
$\frac{1}{f_1} = \frac{2}{f_l}$ Here $2f_l = 18$ cm or $f_l = 9$ cm So, $\frac{1}{9} = \frac{2}{f_l}$ or $f_l = 18$ cm Using, $\frac{1}{f_l} = (\mu - 1)\left(\frac{2}{R}\right)$ or $\frac{1}{18} = (1.5 - 1)\left(\frac{2}{R}\right)$ $\therefore R = 18$ cm when liquid is put between, then $\frac{1}{f_2} = \frac{2}{f_l} + \frac{2}{f}$ or $\frac{1}{(27/2)} = \frac{2}{18} + \frac{2}{f}$ or $f = -54$ cm Now $-\frac{1}{54} = (\mu_1 - 1) \times \frac{1}{R}$ $= (\mu_1 - 1) \times \left(\frac{1}{-18}\right)$ $\therefore \mu_1 = \frac{1}{3} + 1 = \frac{4}{3}$
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