Question

What are the conditions for the minimum deviation by a prism? One face (AC) of a prism of refracting angle 30° and refractive index \( \sqrt{2} \), is silvered. What should be the angle of incidence \( i \) on the surface AB, so that after refraction into the prism, the reflected ray from the silvered surface retraces its path? See the figure.

Hint:

The key for minimum deviation in a prism is that the incident and refracted rays are symmetric. The condition of total internal reflection at the silvered surface is crucial to ensure that the reflected ray retraces its path.

Answer 1

To achieve the condition of minimum deviation in a prism, the following conditions must be met:
1. The incident ray and the refracted ray inside the prism must make equal angles with the prism surface. This means that the angle of incidence at the surface \( AB \) must be equal to the angle of refraction inside the prism.
2. The angle of deviation is minimized when the ray passes symmetrically through the prism. This implies that the path of the light through the prism should be symmetric in nature with respect to the prism's surface.
3. The angle of incidence at the silvered surface (the reflecting surface) must allow total internal reflection. The light should reflect off the silvered surface in such a way that the ray retraces its original path inside the prism. This occurs when the angle of incidence on the surface is greater than the critical angle for total internal reflection.
The relationship between the angles is given by: \[ i + r = A, \] where: - \( i \) is the angle of incidence on surface \( AB \), - \( r \) is the angle of refraction inside the prism, - \( A \) is the refracting angle of the prism. Also, the angle of deviation \( D \) is given by: \[ D = i + r - A. \] For minimum deviation, the deviation is at its smallest value when the ray enters and exits the prism symmetrically. This means that the angle of incidence and the angle of refraction must be equal, i.e., \( i = r \). Thus, for minimum deviation: \[ D_{\text{min}} = 2i - A. \] Now, considering the silvered surface at the other end of the prism, the ray must undergo total internal reflection. The critical angle \( \theta_c \) for total internal reflection is given by: \[ \sin \theta_c = \frac{1}{n}, \] where \( n = \sqrt{2} \) is the refractive index of the prism. Therefore, the critical angle is: \[ \theta_c = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) = 45^\circ. \] For total internal reflection to occur, the angle of incidence at the silvered surface must be greater than this critical angle. Thus, we set the angle of incidence \( i \) such that the reflected ray retraces its path. The angle \( i \) must satisfy: \[ i = 45^\circ. \] Thus, the angle of incidence \( i \) on the surface \( AB \) is 45° for the reflected ray to retrace its path after total internal reflection at the silvered surface.