 # water(with refractive index) = 4/3 in a tank is 18cm deep . oil of refractive index 7/4 lies on water making a convex surface of radius of curvature R = 6 cm as shown , consider oil to act as a this 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

10 years ago

Dear Sanjay,

he law of distances in the case of refraction at an interface between two media of refractive indices n1and n2 is

n2/v – n1/u = (n2 – n1)/R where ‘v’ is the imagedistance, ‘u’ is the object distance and ‘R’ is the radius of curvature of the refracting surface. In the case of the refraction at the air-oil interface we have n2 = 7/4, n1 = 1, u = – 24 cm and R = 6 cm. so that

7/4v – 1/(– 24) = [(7/4) –1]/6

Therefore, 7/4v = 3/24 – 1/24 = 1/12

This gives v = 21 cm.

The image formed (at 21 cm from the oil surface) by the refraction at the air-oil interface will act as a virtual object for the refraction at the oil-water interface so that u = 21 cm (positive according to Cartesian sign convention). In this case n2 = 4/3, n1 = 7/4 and R = ∞.

[The radius of curvature is infinity since the oil-water interface is plane].

Therefore we have

4/3v1 – 7/(4×21) = [(4/3) – (7/4)]/∞ where v1 is the distance of the final image (from the water surface).

Or, 4/3v1 – 7/(4×21) = 0

This gives v1 = 16 cm.

The water column in the tank has a height of 18 cm. Therefore the final image is 2 cm above the bottom of the tank.

Therefore, x = 2

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