Question

A monochromatic ray of light is incident normally on a thin prism of refracting angle A. The ray is deviated through an angle $(1.15)^\circ$ in passing through the prism. The ray reflected internally from the second face emerges from the first face making an angle of $(6.3)^\circ$ with the incident ray. The refractive index of the prism is ______.

• A. 1.625
• B. 1.575
• C. 1.525
• D. 1.515

Hint:

For a thin prism with normal incidence, the angle it hits the second face is always $A$. If it reflects back, it hits the first face at $2A$ and emerges at $2\mu A$. Setting up these two simple linear equations solves the problem instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We are given two distinct optical paths for a ray hitting a thin prism. The first path is standard transmission through the prism, and the second involves an internal reflection. We must use the deviation angles to find the refractive index ($\mu$).

Step 2: Detailed Explanation:
Let the refracting angle of the thin prism be $A$, and its refractive index be $\mu$.
Since the ray is incident normally on the first face, the angle of incidence $i_1 = 0^\circ$, meaning the angle of refraction at the first face $r_1 = 0^\circ$.
Case 1: Direct Transmission
For a thin prism, the total deviation $\delta$ is strictly given by:
$\delta = (\mu - 1)A$
We are given $\delta = 1.15^\circ$.
$(\mu - 1)A = 1.15$ --- (Equation 1)
Case 2: Internal Reflection
The ray hits the second face at an angle of incidence equal to $A$ (since $r_1 + r_2 = A \implies 0 + r_2 = A \implies r_2 = A$).
Instead of emerging, the ray is reflected internally. By the law of reflection, it bounces back at an angle $A$.
It travels back to the first face. The angle it strikes the first face from the inside is geometrically $2A$.
Now, it refracts out of the first face into the air. Using Snell's law for small angles ($\sin \theta \approx \theta$):
$\mu (2A) = 1 \cdot e'$
The angle of emergence is $e' = 2\mu A$.
The problem states this emerging ray makes an angle of $6.3^\circ$ with the original incident ray. Since the incident ray was perfectly normal (perpendicular) to the first face, the angle it makes with the incident ray is exactly the angle of emergence $e'$.
$e' = 6.3^\circ$
$2\mu A = 6.3$
$\mu A = 3.15$ --- (Equation 2)
Solving the Equations:
Expand Equation 1:
$\mu A - A = 1.15$
Substitute $\mu A = 3.15$ from Equation 2 into the expanded Equation 1:
$3.15 - A = 1.15$
$A = 3.15 - 1.15 = 2.0^\circ$
Now substitute $A = 2.0$ back into Equation 2 to find $\mu$:
$\mu (2.0) = 3.15$
$\mu = \frac{3.15}{2.0} = 1.575$

Step 3: Final Answer:
The refractive index is 1.575, matching option (b).