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.
.

Answer

B

Explanation

2 For the covex spherical refracting surface of oil we apply
`(-mu_1)/(u)+(mu_2)/(v)=(mu_2-mu_1)/(R)`
`:'``(-1)/((-24))+((7)/(4))/(v)=((7)/(4)-1)/(6)`
`:'` `v=21cm`
For water oil interface
`((-7)/(4))/(21)+((4)/(3))/V^(`)=0`
`:'` `V^(`)=16cm.`
This is the image distance from water-oil interface. Therefore the distance of the image from the bottom of the tank is 2cm.