Water (with refractive index \(=\frac{4}{3}\) ) in a tank is 18 cm deep. Oil of refractive index \(\frac{7}{4}\) lies on water making a convex surface of radius of curvature ' \(\mathrm{R}=6 \mathrm{~cm}\) ' 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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Geometrical Optics

Water (with refractive index

= \frac{4}{3}

) in a tank is 18 cm deep. Oil of refractive index

\frac{7}{4}

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

Answer: 2

Solution

Refraction at oil surface:

\frac{n _{2}}{v _{1}} - \frac{n _{1}}{u} = \frac{n _{2} - n _{1}}{R}

. With

n_{1} = 1

n_{2} = 7/4

u = - 24

cm,

R = 6

cm:

\frac{7/4}{v _{1}} + \frac{1}{24} = \frac{3/4}{6}

, giving

v_{1} = 21

cm below oil surface. Refraction at oil-water interface:

\frac{4/3}{v _{2}} - \frac{7/4}{21} = 0

, giving

v_{2} = 16

cm below water surface. Distance from bottom:

x = 18 - 16 = 2

cm.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Integer answer

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JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Geometrical Optics

Solution

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