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A convex lens (f = 20 cm, $μ$ = 2) is dipped half in liquid as shown. The two images are separated by 30 cm when an object O is at 60 cm from the lens. The refractive index of liquid is

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1.2

1.33

1.5

1.43

answer is A.

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Detailed Solution

$\frac{1}{f} = \frac{1}{v} + \frac{1}{u} \Rightarrow \frac{1}{20} = \frac{1}{60} + \frac{1}{v} \Rightarrow v = 30 cm \frac{1}{f} = (μ - 1) (\frac{1}{R_{1}} + \frac{1}{R_{2}}) for air part = (\frac{1}{R_{1}} + \frac{1}{R_{2}}) since μ = 2$

$\frac{1}{f^{r}} = \frac{μ_{2} - 1}{f} for water part.$

$Here v = 30 + 30 = 60 cm$

$∴ \frac{μ_{2} - 1}{f} = \frac{μ_{2} - 1}{20} = \frac{1}{60} + \frac{1}{60} = \frac{1}{30}$

$μ_{2} - 1 = \frac{20}{30} \Rightarrow μ_{2} = \frac{5}{3} = \frac{μ_{g}}{μ_{ℓ}} = \frac{2}{μ_{ℓ}}$

$∴ μ_{ℓ} = \frac{6}{5} = 1.2$

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