JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Geometrical Optics

The image of an object, formed by a plano-convex lens at a distance of

8 m

behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is

\frac{2}{3}

times the wavelength in free space. The radius of the curved surface of the lens is

Answer: (C)

Solution

Given: image distance

v = 8 m

, magnification

m = - \frac{1}{3}

.

Using

m = \frac{f - v}{f}

- \frac{1}{3} = \frac{f - 8}{f}

- \frac{f}{3} = f - 8

\frac{4 f}{3} = 8

f = 6 m

Since

λ^{'} = \frac{2}{3} λ_{0}

and

μ \propto \frac{1}{λ}

μ = \frac{λ _{0}}{λ ^{'}} = \frac{λ _{0}}{\frac{2}{3} λ _{0}} = \frac{3}{2}

For a plano-convex lens:

\frac{1}{f} = (μ - 1) (\frac{1}{R} - \frac{1}{\infty}) = (μ - 1) \frac{1}{R}

\frac{1}{6} = (\frac{3}{2} - 1) \frac{1}{R} = \frac{1}{2 R}

R = 3 m

Hence the answer is (C).

JEE Advanced · 2013 · 02 Jun 2013 · Paper 1 · Single correct option