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

`2`

Explanation

First refraction:`mu_(1)=1,u=-24,mu_(2)=(7)/(4)`
`R=+6,(mu_(2))/(v)-(mu_(1))/(u)=(mu_(2)-mu_(1))/(R )`
After solving `v=21`
Now for second refraction:
`h=(21)/((21//16))=16`
So from bottoms`18-16=2`
So,`x=2`