Question

A real object is kept in air and single refraction occurs at the convex boundary of a spherical glass surface. For the image to be real, the real object distance $({\mu }_g=\frac{3}{2})$

• A. Can be greater than three times the radius of curvature of the refracting surface
• B. Should be greater than two times the radius of curvature of the refracting surface
• C. Should be greater than the radius of curvature of the refracting surface
• D. Is independent of the radius of curvature of the refracting surface

Answer 1

Answer: (D)

Is independent of the radius of curvature of the refracting surface

By sign convention, in this case, $u$ is negative and $R$ is positive.
$\frac{{\mu }_g}{v}=\frac{-1}{u}+\frac{{\mu }_g-1}{R}$
For image to be real, $v$ should be positive as can be seen from the figure.
In the above equation, $\frac{-1}{u}$ is positive as $u$ is negative and $\frac{{\mu }_g-1}{R}=\frac{0.5}{R}$ is positive as $R$ is positive.
Thus $\frac{1}{v}$ and hence $v$ will be always positive
So, image of the real object will be real for any object distance irrespective of the radius of curvature of the refracting surface.
So (A) and (D) are correct.