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A denser medium of refractive index 1.5 has a concave surface of radius of curvature 12cm. An object is in denser medium at a distance of 9cm, from the pole. Then the image is virtual at a distance ____

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4.8cm

2.4cm

9.6cm

7.2cm

answer is A.

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Detailed Solution

$\frac{μ_{2}}{u} + \frac{μ_{1}}{v} = \frac{μ_{2} - μ_{1}}{- R}$

$\frac{1.5}{9} + \frac{1}{v} = \frac{1.5 - 1}{- 12} \Rightarrow V = - 4.8 cm$

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