Water (with refractive index=4/3) in a tank is 18 cm deep. Oil of refractive index 7/4 lies on water making a convex surface of radius of curvature 'R=6cm' as shown. 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 :

Here, the usual lens equation won't be applicable as the medium is different on two sides of the lens. So we will consider refraction at two surfaces. For first refraction, (by sign convention) μ₁=1, μ₂=7/4, u=-24cm, R=6cm Putting this into the formula, we get 7/4v - 1/-24 = 7/4 - 1/6 Solving this, we get v=21cm As the lens is thin, this can be taken as the object distance for the second refraction. For the second refraction, (by sign convention) μ₁=7/4, μ₂=4/3, u=21cm, R=∞ (plane surface) Putting this into the formula, we get 4/3v - 7/4(21) = 4/3 - 7/4(1/∞) Solving this, we get v=16cm i.e. it is 16 cm below the water level. The tank is 18cm deep. So the height of the image from the bottom of the tank is 18-16=2cm