A biconvex lens of refractive index 1.5 has a focal length of 20 cm in air. Its focal length when immersed in a liquid of refractive index 1.6 will be: (a) - 16 cm (b) +16 cm (c) +160 cm (d) - 160 cm

Answer: (d). (fl)/(fa) = (μ - 1)/((μ)/(μl) - 1) (μ)/(μl) = (1.5)/(1.6) = 0.9375, so fl/20 = (0.5)/(-0.0625) = -8 fl = -160 cm (the lens becomes diverging) So the answer is (d) -160 cm.

A biconvex lens of refractive index 1.5 has a focal length of 20 cm… | JEE Main 2024 (29 Jan, Shift 1)

Q39JEE Main 2024 · 29 JAN · SHIFT 1Easy

Spherical Surfaces & Thin Lenses

A biconvex lens of refractive index 1.5 has a focal length of $20\text{ cm}$ in air. Its focal length when immersed in a liquid of refractive index 1.6 will be:

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Answer: (d)

$\frac{f_l}{f_a} = \frac{\mu - 1}{\frac{\mu}{\mu_l} - 1}$
$\frac{\mu}{\mu_l} = \frac{1.5}{1.6} = 0.9375$ , so $\frac{f_l}{20} = \frac{0.5}{-0.0625} = -8$
$f_l = -160\text{ cm}$ (the lens becomes diverging)
So the answer is (d) $-160$ cm.

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