Question

A spherical surface of radius of curvature R, separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O and PO = O The distance PO is equal to

• A. 5 R
• B. 3 R
• C. 2 R
• D. 1.5 R

Answer 1

The Correct Option is A

The correct option is(A): 5 R.

Let us say PO = OQ = X 
Applying\(\, \, \, \, \, \, \, \, \frac{\mu_2}{v}-\frac{\mu_1}{u}=\frac{\mu_1-\mu_2}{R}\) 
Substituting the values with sign 
\(\frac{1.5}{+X}-\frac{1.0}{-X}=\frac{1.5-1.0}{+R}\) 
(Distances are measured from O and are taken as positive in 
the direction of ray of light) 
\(\therefore\hspace15mm \frac{2.5}{X}=\frac{0.5}{R}\)
\(\therefore\hspace15mm X=5R\)