Water (with refractive index = 4/3) in a tank is `18 cm` deep. Oil of refraction index `7//4` lies on water making a convex surface of radius of curvature `R = 6 cm` as shown in Fig. Consider oil to act as a thin lens. An object `S` is placed `24 cm` above water surface. The location of its image is at `x cm` above the bottom of the tank. Then `x` is.
.

Options

A. `1`
B. `2`
C. `3`
D. `4`

Answer

B

Explanation

For refraction at air-oil interface, we have
`u = - 24 cm, mu_(1) = 1, mu_(2) = (7)/(4)`,
`R = + 6 cm, v = v_(1)` (say)
As `-(nu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`:. -(1)/(-24) + (7//4)/(v_(1)) = ((7//4) - 1)/(6) = (3)/(24)`
or `(7)/(4 v_(1)) = (3)/(24) - (1)/(24) = (2)/(24) = (1)/(12)`
or `v_(1) = (12 xx 7)/(4) = 21 cm`.
This image will act as object for oil water interface.
Taking refraction at oil-water interface, we have
`u = + 21 cm, v = v_(2), mu_(1) = (7)/(4), mu_(2) = (4)/(3), R = oo`
As `-(mu_(1))/(u) + (mu_(2))/(v) = (mu_(2) - mu_(1))/(R )`
`:. (-(7//4))/(21) + ((4//3))/(v_(2)) = (4//3 - 7//4)/(oo) = 0`
or `v_(2) = 16 cm`
Hence, `x = 18 - 16 = 2 cm`