Water (with refractive index ) in a tank is 18 cm deep. Oil of refractive index lies on water making a convex surface of radius of curvature as shown. Consider oil to act as a thin lens. An object is placed 24 cm above the water surface. The location o

Question

Water (with refractive index $= \frac{4}{3}$ ) in a tank is 18 cm deep. Oil of refractive index $\frac{7}{4}$ lies on water making a convex surface of radius of curvature $' R = 6 cm'$ as shown. Consider oil to act as a thin lens. An object $' S'$ is placed 24 cm above the water surface. The location of its image is at $' x'$ cm above the bottom of the tank. Then $' x'$ is _____ [2011]

Smart Achievers · Accepted Answer

(2)

For the convex spherical refracting surface, i.e., air-oil interface. Since oil is acting as a thin lens, we will ignore its dimension.

$\frac{7}{4}$

$- \frac{μ_{1}}{u} + \frac{μ_{2}}{v} = \frac{μ_{2} - μ_{1}}{R}$

$∴$ $\frac{- 1}{(- 24)} + \frac{\frac{7}{4}}{v} = \frac{\frac{7}{4} - 1}{6}$

$∴$ $v = 21 cm$

This image will act as object for the water-oil interface.

$u = 21 cm, v = v', μ_{1} = \frac{7}{4}, μ_{2} = \frac{4}{3}, R = \infty$

$\frac{\frac{- 7}{4}}{21} + \frac{\frac{4}{3}}{v'} = 0$

$∴$ $v' = 16 cm$

Therefore, the distance of the image from the bottom of the tank is $18 - 16 = 2 cm$

Solution

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