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.
Answer & explanation
Correct answer: option 2
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.
$⇒x+x=v$
$⇒x=\frac{v}{2}=\frac{3R}{2}$
$⇒x=1.5R$