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# Plane surface of plane - convex lens is silvered and it then acts like a concave mirror of focal length . 30 cm. If the refractive index of the lens is 1.5 , the radius of curvature of convex surface is

$(a)\;15\;cm \\ (b)\;30\;cm \\ (c)\;60\;cm \\ (d)\;90\;cm$

Now system consists of a plane convex lens and a plane mirror and light incident on it will under go 2 refractions and one reflection.
$\therefore$ the effective focal length is
$\large\frac{1}{F} =\frac{2}{f} +\frac{1}{f_m}$
$\large\frac{1}{F} =\frac{2}{f}$
$[f_m=\infty , f$ -focal length of lens]
F is given to be 30 cm
$\therefore \large\frac{1}{30} =\frac{2}{f}$
$\therefore f=60 \;cm$
We know $\large\frac{1}{f} =(\mu-1) \bigg[ \large\frac{1}{R_1} -\frac{1}{R_2}\bigg]$
$\large\frac{1}{60}$$=(1.5 -1) \bigg[ \large\frac{1}{R_1} -\frac{1}{\infty}]$
$\large\frac{1}{60} =\frac{0.5}{R_1}$
the radius of curvature of lens.
$R_1=30 cm$
Hence b is the correct answer.

edited Jul 15, 2014