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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
\( \mathrm{P} \) \( O A=18 \mathrm{~cm} \), its real inverted image is formed at \( A \) itself, as

W shown in figure. When a liquid of refractive index \( \mu_{l} \) is put between the lens and the mirror, the pin has to be moved to \( A^{\prime} \), such that \( O A^{\prime}=27 \mathrm{~cm} \), to get its inverted real image at \( A^{\prime} \) itself. The value of \( \mu_{l} \) will be
[JEE (Main) 2019]
(a) \( \frac{4}{3} \)
(b) \( \sqrt{2} \)
(c) \( \sqrt{3} \)
(d) \( \frac{3}{2} \)
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