A spherical surface of radius of curvature R separates air from glass (\( \mu=1.5 \)). 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 = OQ. The distance PO is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 5R \)
For refraction at a spherical surface, \( \displaystyle \frac{n_2}{v}-\frac{n_1}{u}=\frac{n_2-n_1}{R} \). All distances are measured from the pole O using the Cartesian sign convention, with the incident light travelling in the positive direction.
Let PO = OQ = \( x \). The light goes from air (\( n_1=1 \)) into glass (\( n_2=1.5 \)). The object P is in front of the surface, so \( u=-x \). The real image Q is in the glass, so \( v=+x \). Also, \( R \) is positive because the centre of curvature is in the glass.
\( \displaystyle \frac{1.5}{x}-\frac{1}{-x}=\frac{1.5-1}{R} \)
\( \displaystyle \frac{2.5}{x}=\frac{0.5}{R} \)
\( \Rightarrow\; x=5R \)
Therefore, PO = \( 5R \).