Explanation

(b) When an object is placed in front of such a lens, the rays are first of all refracted from the convex surface and again refracted from convex surface.
Let $f₁,f_{m}$ be focal lengths of convex surface and mirror (plane polished surface) respectively, then effective focal length is
$\frac{1}{F} = \frac{1}{f₁} + \frac{1}{f_{m}} + \frac{1}{f₁} = \frac{2}{f₁} + \frac{1}{f_{m}}$
Since,
$f_{m} = \frac{R}{2} = \ \infty$
$\therefore\frac{1}{F} = \frac{2}{f₁}$
From lens formula
$\frac{1}{f₁} = (\mu - 1)\left( \frac{1}{R} \right)$
$\therefore\frac{1}{F} = \frac{2(\mu - 1)}{R}$
$\Longrightarrow F = \frac{R}{2(\mu - 1)}$
$or\ \ R_{eq} = 2F = \frac{R}{(\mu - 1)}$