Question

A ray of light incidents at an angle of $9.3^\circ$ on one face of a small angle prism of refracting angle $6^\circ$. If the ray of light emerges normally from the second face, the refractive index of the material of the prism is

• A. 1.40
• B. 1.45
• C. 1.55
• D. 1.50

Hint:

For a prism, the key relation is $r_1 + r_2 = A$. Normal emergence means the angle of emergence $e=0$ and the angle of refraction at the second face $r_2=0$. This immediately simplifies the relation to $r_1=A$.

Answer 1

The Correct Option is C

Step 1: State the relations for a small-angle prism.
For a small-angle prism, the angles of incidence ($i$), refraction ($r_1$), emergence ($e$), and the refracting angle ($A$) are related by:
1. $r_1 + r_2 = A$
2. Snell's Law at the first face: $\sin i = \mu \sin r_1$. For small angles, $i \approx \mu r_1$.
3. Snell's Law at the second face: $\sin e = \mu \sin r_2$. For normal emergence, $e=0$.

Step 2: Apply the condition for normal emergence.
If the ray emerges normally from the second face, the angle of emergence $e$ is $0^\circ$, and the angle of refraction at the second face, $r_2$, is also $0^\circ$.
$r_2 = 0^\circ$.

Step 3: Find the angle of refraction at the first face.
Using the prism relation $r_1 + r_2 = A$:
$r_1 + 0^\circ = A \implies r_1 = A$.
We are given the refracting angle $A=6^\circ$. So, $r_1 = 6^\circ$.

Step 4: Use Snell's Law at the first face to find the refractive index $\mu$.
Snell's Law at the first face (incidence $i$) is $\sin i = \mu \sin r_1$.
We are given $i = 9.3^\circ$ and we found $r_1 = 6^\circ$.
\[ \mu = \frac{\sin i}{\sin r_1} = \frac{\sin(9.3^\circ)}{\sin(6^\circ)}. \]

Step 5: Calculate the numerical value.
We use the small angle approximation $\sin\theta \approx \theta$ (in radians) for the ratio of the angles: \[ \mu \approx \frac{i}{r_1} = \frac{9.3^\circ}{6^\circ} = 1.55. \] (The approximation is accurate for this ratio).
The refractive index is $\mu = 1.55$.