A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. The refractive index $n$ of the first lens is 1.5 and that of the second lens is 1.2. Both the curved surfaces have 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 [2012]

Question

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(2)

The focal length $R$ of the plano-convex lens with $n = 1.5$ using the lens-maker formula is

$\frac{1}{f_{1}} = (n_{1} - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}] = (1.5 - 1) [\frac{1}{14} - \frac{1}{\infty}] = \frac{1}{28}$

The focal length $(f_{2})$ of the plano-convex lens with $n = 1.2$ is

$\frac{1}{f_{2}} = (n_{2} - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}] = (1.2 - 1) [\frac{1}{\infty} - \frac{1}{- 14}] = \frac{1}{70}$

Focal length $F$ of the combination,

$\frac{1}{F} = \frac{1}{f_{1}} + \frac{1}{f_{2}} = \frac{1}{20}$

Now, applying the lens formula for the combination of lenses,

$\frac{1}{V} - \frac{1}{U} = \frac{1}{F}$ $\Rightarrow \frac{1}{V} - \frac{1}{- 40} = \frac{1}{20}$ $[Given u = 40 cm]$

$∴$ $V = 40 cm$

Solution

1 − 280.0 cm

2 40.0 cm

3 21.5 cm

4 13.3 cm

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