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 $P O = O Q$ . The distance PO is equal to

• A. 5 R
• B. 3 R
• C. 2 R
• D. 1.5 R

Answer 1

Answer: (A)

5 R

Answer 2

By using $\frac { \mu _ { 2 } } { v } - \frac { \mu _ { 1 } } { u } = \frac { \mu _ { 2 } - \mu _ { 1 } } { R }$

Where $\mu _ { 1 } = 1$ $\mu _ { 2 } = 1.5$ u = – OP, v = OQ

Hence $\frac { 1.5 } { O Q } - \frac { 1 } { - O P } = \frac { 1.5 - 1 } { ( + R ) }$ ⇒ $\frac { 1.5 } { O P } + \frac { 1 } { O P } = \frac { 0.5 } { R }$

⇒ OP = 5 R