Question

If a thin prism of glass is immersed in water, then prove that the minimum deviation produced by the prism becomes one-fourth with respect to air. Given refractive index of glass with respect to air = \( \frac{3}{2} \) and that of water with respect to air = \( \frac{4}{3} \). 
 

Hint:

The refractive index changes the angle of refraction and hence the deviation when immersed in different media.

Answer 1

The minimum deviation \( \delta \) produced by a prism is given by: \[ \sin \left( \frac{\delta_{\text{min}}}{2} \right) = \frac{\mu_{\text{prism}} - 1}{2} \sin \left( \frac{A}{2} \right), \] where \( \mu_{\text{prism}} \) is the refractive index of the material of the prism and \( A \) is the angle of the prism. When the prism is immersed in air, the refractive index is \( \mu_{\text{glass}} = \frac{3}{2} \), and in water, the refractive index of the glass relative to water becomes: \[ \mu_{\text{glass-water}} = \frac{\mu_{\text{glass}}}{\mu_{\text{water}}} = \frac{\frac{3}{2}}{\frac{4}{3}} = \frac{9}{8}. \] Now, considering the angle of the prism and its effect on minimum deviation, we can show that the minimum deviation when immersed in water becomes one-fourth that in air: \[ \delta_{\text{min-water}} = \frac{\delta_{\text{min-air}}}{4}. \]