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

For a spherical surface, the lens-maker's equation is given by: \[ \frac{1}{f} = (n_2 - n_1) \left( \frac{1}{R_1} - \frac{1}{R_2} \right) \] where \( n_1 \) and \( n_2 \) are the refractive indices of the medium on either side of the surface, \( R_1 \) is the radius of curvature of the surface, and \( R_2 \) is the radius of curvature of the second surface. In this case, we are dealing with a spherical convex surface, so \( R_2 = \infty \) (since it is an open surface), and the equation simplifies to: \[ \frac{1}{f} = \left( n_{\text{glass}} - n_{\text{air}} \right) \frac{1}{R} \] Substitute \( n_{\text{glass}} = 1.5 \) and \( n_{\text{air}} = 1 \): \[ \frac{1}{f} = (1.5 - 1) \frac{1}{R} = \frac{0.5}{R} \] Thus, the focal length is: \[ f = \frac{2R}{1} \] Since the object is placed at a distance \( \frac{R}{2} \) from the surface, we can use the lens formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where \( u = -\frac{R}{2} \) (object distance is negative), and \( f = \frac{2R}{1} \). Substituting the values: \[ \frac{1}{\frac{2R}{1}} = \frac{1}{v} - \frac{1}{-\frac{R}{2}} \] Simplifying: \[ \frac{1}{2R} = \frac{1}{v} + \frac{2}{R} \] \[ \frac{1}{v} = \frac{1}{2R} - \frac{2}{R} = -\frac{3}{2R} \] Thus, the image distance is: \[ v = -\frac{2R}{3} \] The negative sign indicates that the image is virtual, formed on the same side as the object. Therefore, the image is virtual, formed at a distance \( \frac{2R}{3} \) behind the surface.