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.
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Answer

2

Explanation

`(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R)`
`(7)/(4v)-(1)/(-24)=(7/4-1)/(6)`
`(7)/(4v)-(1)/(-24)-(1)/(24)=(2)/(24)=(1)/(12)`
` (7xx12)/(4)=V=21 cm`
`(21)/(OS)=(7//4)/(4//3) (21)/(OS)=(7)/(4)xx(3)/(4)`
`OS=16`
`therefore BS=2 cm`