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 `-(mu_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`.