Object O is placed at a distance of 14cm from an equi convex lens $L_{1}$ of refractive index $μ = \frac{3}{2}$ and radius of curvature of lens surface is 32 cm. One of its surface is silvered another convex lens $L_{2}$ of focal length 24 cm is placed between object and silvered lens such that the final image coincides with the object. The distance between object and the lens $L_{2}$ in cm is ……… ( Take first Refraction from Lens $L_{2}$ )

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answer is 6.

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Detailed Solution

$\frac{1}{v} - \frac{1}{- x} = \frac{1}{24} \Rightarrow \frac{1}{v} = \frac{1}{24} - \frac{1}{x} \frac{1}{v} = - \frac{1}{(\frac{24 - x}{24 x})}$

Object distance for silvered lens is $(14 - x) + \frac{24 x}{(24 - x)}$ for image to be an object O, this distance must be equal to equivalent radius of mirror.
For (Reflecting lens is effectively mirror)

$- \frac{2}{R_{e q}} = 2 (\frac{3}{2} - 1) (\frac{1}{32} - \frac{1}{- 32}) - \frac{3}{- 32} \Rightarrow R_{e q} = - 16 c m ∴ 16 = (14 - x) + \frac{24 x}{(24 - x)} \Rightarrow x = 6 c m$