Explanation

(a) A plano-convex lens behaves as a concave mirror if it’s one surface (curved) is silvered. The rays refracted from plane surface are reflected from curved surface and again refract from plane surface. Therefore, in this lens two refractions and one reflection occur.
Let the focal length of silvered lens is$\ F.$
$\frac{1}{F} = \frac{1}{f} + \frac{1}{f} + \frac{1}{f_{m}} = \frac{2}{f} + \frac{1}{f_{m}}$
Where $f = \ $focal length of lens before silvering
$f_{m} =$focal length of spherical mirror.
$\therefore\frac{1}{F} = \frac{2}{f} + \frac{2}{R}\ \ \ \ \ \ \ \ \ \ldots..(i)$
$(\because R = 2f_{m})$
$now\ ,\frac{1}{f} = (\mu - 1)\left( \frac{1}{R₁} - \frac{1}{R₂} \right)\ \ \ldots.(ii)$
$here,\ \ \ R₁ = \ \infty,R₂ = 30\ cm$
$\therefore\ \ \frac{1}{f} = (1.5 - 1)\left( \frac{1}{\infty} - \frac{1}{30} \right)$
$\Longrightarrow \ \frac{1}{f} = \ - \frac{0.5}{30} = - \frac{1}{60}$
$\Longrightarrow f = \ - 60\ cm$
Hence from eq (i)
$\frac{1}{F} = \frac{2}{60} + \frac{2}{30} = \frac{6}{60}$
$F = 10\ cm$
Again given that,
Size of object = size of image
$\Longrightarrow \ O = I$
$\therefore\ m = \ - \frac{v}{u} = \frac{I}{O} \Longrightarrow \frac{v}{u} = \ - 1$
$\Longrightarrow v = - u$
Thus, from lens formula
$\frac{1}{F} = \frac{1}{v} - \frac{1}{u}$
$\frac{1}{10} = \frac{1}{- u} - \frac{1}{u}$
$\frac{1}{10} = \ - \frac{2}{u}$
$\therefore\ u = \ - 20\ cm$
Hence, to get a real image, object must be placed at a distance 20 cm on the left side of lens.