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 $n \times 10 c m$ then n is

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Detailed Solution

Using the lens formula,
$\frac{1}{v} - \frac{1}{u} = \frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} o r \frac{1}{v} = \frac{1}{u} + \frac{1}{f_{1}} + \frac{1}{f_{2}}$
$= \frac{1}{u} + (n_{1} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}}) + (n_{2} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$
Submitting the values, we get
$\frac{1}{v} = \frac{1}{- 40} + (1.5 - 1) (\frac{1}{14} - \frac{1}{\infty}) + (1.2 - 1) (\frac{1}{\infty} - \frac{1}{14})$
Solving this equation, we get
$v = + 40 c m$