Explanation

(a) Focal length of lens
$\frac{1}{f} = (\mu - 1)(\frac{1}{R_{1}} - \frac{1}{R_{2}})$
For equi-convex lens,
$R_{1} = + R,\ R_{2} = \ - R$
$\therefore\ \frac{1}{f} = \ (\mu - 1)\left( \frac{1}{R} - \frac{1}{- R} \right)$
$\frac{1}{f} = (\mu - 1)\left( \frac{2}{R} \right)$
$f = \frac{R}{2(\mu - 1)}$
$f < R,\ so,\ 2(\mu - 1) < 1$
$(\mu - 1) < \frac{1}{2}$
$(\mu - 1) < 0.5$
$\mu < 1.5$
Focal length of convex lens is positive. So, $\mu\ $cannot be negative, hence should be greater than zero but less than 1.5