A double convex lens has power P and same radii of curvature R of both the surfaces.

(4) Given, using lens maker's formula

$\frac{1}{f}=(k-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)$

Here, $R_{1}=R_{2}=R$ (For double convex lens)

$\therefore \frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right)$

$\Rightarrow P=\frac{1}{f}=(\mu-1) \frac{2}{R}$             .....(i)

For plano convex lens,

$R_{1}=R^{\prime}, R_{2}=\infty$

Using lens maker's formula again, we have

$1.5 P=(\mu-1)\left(\frac{1}{R^{\prime}}-\frac{1}{\infty}\right)$                 ....(ii)

$\Rightarrow \frac{3}{2} P=\frac{\mu-1}{R^{\prime}}$

From (i) and (ii),

$\frac{3}{2}=\frac{R^{\prime}}{2 R} \Rightarrow R^{\prime}=\frac{R}{3}$