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A plano-convex lens is made of a material of refractive index n. When a small object is placed 30 cm away in front of the curved surface of the lens, an image of double the size of the object is produced. Due to reflection from the convex surface of the lens, another faint image is observed at a distance of 10 cm away from the lens. Which of the following statement(s) is (are) true?

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The refractive index of the lens is 2.5

The radius of curvature of the convex surface is 45 cm

The faint image is erect and real

The focal length of the lens is 20 cm

answer is A, D.

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Detailed Solution

For lens applying lens formula

$\frac{1}{v} - \frac{1}{u} = \frac{1}{f} \Rightarrow \frac{1}{v} - \frac{1}{- 30} = \frac{1}{f}$

$\frac{1}{60} - \frac{1}{- 30} = \frac{1}{f} \Rightarrow \frac{1}{f} = \frac{1}{60} + \frac{1}{30}$

$∴ f = 20 c m$

$∴ m = \frac{v}{u} \Rightarrow - 2 = \frac{v}{- 30} \Rightarrow v = 60 c m$

For reflection, applying lens formula,

$\frac{1}{v} + \frac{1}{u} = \frac{1}{f} = \frac{2}{R} \Rightarrow \frac{1}{10} + \frac{1}{- 30} = \frac{2}{R} \Rightarrow R = 30 c m$

The image formed by convex side is faint erect and virtual.

By lens maker formula

$\frac{1}{f} = (\frac{n_{l}}{n_{s}} - 1) (\frac{1}{R_{1}} - \frac{1}{R_{2}})$

$∴ \frac{1}{20} = (\frac{n_{l}}{1} - 1) (\frac{1}{30}) ∴ n_{l} = 2.5$

(refractive index of lens)

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