Question

A ray is incident at an angle of incidence $i$ on one surface of a small angle prism(with angle of prism A) and emerges normally from the opposite surface. If the refractive index of the material of the prism is $\mu$. then the angle of incidence is nearly equal to:

• A. $\frac {A}{2\mu}$
• B. $\frac {2A}{\mu}$
• C. $\mu A$
• D. $\frac {\mu \,A}{2}$

Answer 1

The Correct Option is C

The problem involves analyzing the refraction of light through a prism. We are given that a ray is incident on one surface of a prism and emerges normally from the opposite surface. Our task is to find the angle of incidence \(i\). Here’s the step-by-step solution:

1. Since the ray emerges normally from the opposite face, the angle of refraction at the second surface is zero.
2. Applying Snell's law at this surface:

$ \mu \sin(r_{2}) = 1 \cdot \sin(0) = 0, \text{ (as the ray emerges normally)} $

1. Since \(r_{2}=0\), there is no deviation at this surface.
2. Now consider the prism. Using the prism formula for small angles, \(A = r_{1} + r_{2}\). Since \(r_{2} = 0\), we have:

$ r_{1} = A $

1. Apply Snell's law at the first surface (incident surface):

$ \sin(i) = \mu \sin(r_{1}) $

1. Substitute \(r_{1}=A\):

$ \sin(i) = \mu \sin(A) $

1. For small angles, \(\sin(A) \approx A\), hence:

$ \sin(i) \approx \mu A $

1. When \(\sin(i)\) is small and approximately equal to the angle itself, \(i \approx \mu A\).
2. Therefore, the angle of incidence is approximately equal to \(i \approx \mu A\).

Thus, the angle of incidence is nearly equal to the product of the refractive index of the material and the prism angle. Therefore, the correct answer is:

\(\mu A\)