Question

A ray of light is incident at an angle 'i' on one face of thin prism. The ray emerges normally from the other face. Refractive index of the glass prism is 'n' and angle of prism is 'A'. The value of 'i' is

• A. $An$
• B. $An^2$
• C. $A/n$
• D. $A/n^2$

Hint:

For thin prisms, always use the small-angle approximation $\sin \theta \approx \theta$ (in radians) to simplify Snell's law and geometric relations significantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We have a thin prism where light enters at angle $i$ and exits normally. We need to express $i$ in terms of the refractive index $n$ and the prism angle $A$.

Step 2: Key Formula or Approach:
1. For a thin prism, $A = r_1 + r_2$.
2. Normal emergence implies the angle of emergence $e = 0$, thus $r_2 = 0$.
3. Snell's Law: $\mu = \frac{\sin i}{\sin r_1}$.

Step 3: Detailed Explanation:
4. Since $r_2 = 0$, the formula $A = r_1 + r_2$ simplifies to $A = r_1$.
5. Applying Snell's Law at the first surface: $n = \frac{\sin i}{\sin r_1}$.
6. For a thin prism, angles $i$ and $r_1$ are small, so $\sin i \approx i$ and $\sin r_1 \approx r_1$.
7. Therefore, $n \approx \frac{i}{r_1} \implies i = n \cdot r_1$.
8. Substituting $r_1 = A$, we get $i = nA$.

Step 4: Final Answer:
The value of $i$ is $An$, which is option (A).