A thin convex lens made from crown glass (μ = 3/2) has focal length f. When it is measured in two different liquids having refractive indices 4/3 and 5/3, it has the focal lengths f₁ and f₂ respectively. The correct relation between the focal lengths is: (a) f₁ = f₂ f and f₂ becomes negative (c) f₂ > f and f₁ becomes negative (d) f₁ and f₂ both become negative

Answer: (b). fl/f = (μ - 1)/((μ)/(μl) - 1) with μ = 1.5. μl = 4/3: (μ)/(μl) = 9/8, f₁/f = (0.5)/(0.125) = 4 ⇒ f₁ = 4f > f μl = 5/3: (μ)/(μl) = 9/10, f₂/f = (0.5)/(-0.1) = -5 ⇒ f₂ = -5f (negative) So the answer is (b).

A thin convex lens made from crown glass (μ = 3/2) has focal length… | JEE Main 2014

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Spherical Surfaces & Thin Lenses

A thin convex lens made from crown glass $\left(\mu = \frac{3}{2}\right)$ has focal length $f$ . When it is measured in two different liquids having refractive indices $\frac{4}{3}$ and $\frac{5}{3},$ it has the focal lengths $f_1$ and $f_2$ respectively. The correct relation between the focal lengths is:

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Answer: (b)

$\frac{f_l}{f} = \frac{\mu - 1}{\frac{\mu}{\mu_l} - 1}$ with $\mu = 1.5$ .
$\mu_l = \frac{4}{3}$ : $\frac{\mu}{\mu_l} = \frac{9}{8}$ , $\frac{f_1}{f} = \frac{0.5}{0.125} = 4 \Rightarrow f_1 = 4f > f$
$\mu_l = \frac{5}{3}$ : $\frac{\mu}{\mu_l} = \frac{9}{10}$ , $\frac{f_2}{f} = \frac{0.5}{-0.1} = -5 \Rightarrow f_2 = -5f$ (negative)
So the answer is (b).

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