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

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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.

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