IIT-JEE 2011 · Paper 2 · Physics Q36

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PhysicsInteger+4Question 36 of 60

Water (with refractive index $=\frac43$ ) in a tank is 18 cm deep. Oil of refractive index $\frac74$ lies on water making a convex surface of radius of curvature ' $R=6$ cm' 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

Tank with water (μ = 4/3) and a thin layer of oil (μ = 7/4) on top whose upper surface is convex with R = 6 cm; object S lies on the axis in air (μ = 1) above.

Your answer

Answer and worked solution

Official answer: 2

Curved oil surface (air → oil, $R=+6$ cm, object 24 cm above):
$\frac{7/4}{v}-\frac1{-24}=\frac{7/4-1}6\implies\frac7{4v}=\frac18-\frac1{24}=\frac1{12}\implies v=21\text{ cm}.$

Thin oil layer onto water (plane surface at the water level):
$\frac{4/3}{v'}=\frac{7/4}{21}=\frac1{12}\implies v'=16\text{ cm below the surface}.$

Height above the bottom. $18-16=2$ cm.

Answer: 2

Topics:Refraction at spherical surfaces Refraction at plane surfaces

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IIT-JEE 2011 · Paper 2, Physics question 36