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 :

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1 m

2 m

3 m

6 m

answer is C.

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Detailed Solution

Refractive index of lens,

$μ = \frac{λ_{vacuum}}{λ_{lens}} = \frac{λ_{vacuum}}{\frac{2}{3} λ_{ivcuam}} = \frac{3}{2}$

Magnification of image,

$\begin{array}{l} m = \frac{h_{i}}{h_{0}} = \frac{- v}{u} \Rightarrow \frac{h_{o} / 3}{h_{o}} = \frac{- v}{u} \Rightarrow v \\ = \frac{- u}{3} \end{array}$

Using lens maker's formula

$\begin{array}{l} \frac{1}{v} - \frac{1}{u} = (μ - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) \\ \frac{1}{v} - \frac{1}{- 3 v} = (\frac{3}{2} - 1) (\frac{1}{\infty} - \frac{1}{R}) \\ \frac{1}{8} - \frac{1}{- 24} = (\frac{3}{2} - 1) (\frac{1}{\infty} - \frac{1}{R}) \\ \frac{1}{6} = \frac{1}{2} \times (- \frac{1}{R}) \end{array}$