Shown in the figure is a spherical surface of radius of curvature R & R.I.(= n)1.5. The distance of the silvering of the plane surface so as to form an image at the pole due to a very far object.

1.5R

Since light comes from very far object

$u=∞$

$⇒\frac{n}{v}-\frac{1}{∞}=\frac{n-1}{R}$

$⇒v=\frac{nR}{n-1}=\frac{1.5R}{1.5-1}=3R$

Due to presence of the plane mirror, the image found at I behaves as a virtual object for the plane mirror & a real image I’ is formed in front of the plane mirror, at the pole P.

$⇒x+x=v$

$⇒x=\frac{v}{2}=\frac{3R}{2}$

$⇒x=1.5R$