A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index n of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces are of the same radius of curvature R=14 cm. For this bi-convex lens, for an object distance of 40 cm, the image distance will be (in cm):

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answer is 40.

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Detailed Solution

$\begin{array}{l} \frac{1}{f} = \frac{1}{f_{1}} + \frac{1}{f_{2}} \\ (μ_{1} - 1) [\frac{1}{R} - \frac{1}{\infty}] + (μ_{2} - 1) [\frac{1}{\infty} - \frac{1}{- R}] \\ = \frac{(1.5 - 1)}{R} + \frac{(1.2 - 1)}{R} \Rightarrow f = 20 c m \\ \frac{1}{ν} - \frac{1}{u} = \frac{1}{f}; u = - 40 \\ \frac{1}{ν} - \frac{1}{- 40} = \frac{1}{20} \Rightarrow ν = 40 c m \end{array}$