Question

A spherical convex surface of radius of curvature \( R \) separates glass (refractive index \( 1.5 \)) from air. Light from a point source placed in air at a distance \( \frac{R}{2} \) from the surface falls on it. Find the position and nature of the image formed.

Hint:

For a convex spherical surface, the image is virtual and formed on the same side as the object if the object is placed closer than the focal point.

Answer 1

The lens-maker's equation for a spherical surface is \( \frac{1}{f} = (n_2 - n_1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \), where \( n_1 \) and \( n_2 \) are the refractive indices of the surrounding media, and \( R_1 \) and \( R_2 \) are the radii of curvature of the surfaces. For a spherical convex surface, \( R_2 = \infty \), simplifying the equation to \( \frac{1}{f} = \left( n_{\text{glass}} - n_{\text{air}} \right) \frac{1}{R} \). With \( n_{\text{glass}} = 1.5 \) and \( n_{\text{air}} = 1 \), this becomes \( \frac{1}{f} = (1.5 - 1) \frac{1}{R} = \frac{0.5}{R} \). Consequently, the focal length is \( f = \frac{2R}{1} \). Using the lens formula \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \) with the object placed at \( u = -\frac{R}{2} \) and \( f = \frac{2R}{1} \): \( \frac{1}{\frac{2R}{1}} = \frac{1}{v} - \frac{1}{-\frac{R}{2}} \). This simplifies to \( \frac{1}{2R} = \frac{1}{v} + \frac{2}{R} \). Rearranging for \( \frac{1}{v} \): \( \frac{1}{v} = \frac{1}{2R} - \frac{2}{R} = -\frac{3}{2R} \). Thus, the image distance is \( v = -\frac{2R}{3} \). The negative sign indicates a virtual image formed on the same side as the object, at a distance of \( \frac{2R}{3} \) behind the surface.