Question

Light is incident at such an angle so that minimum deviation takes place. Now a film of refractive index RI = \(n/2\) is stick on other face such that total internal reflection takes place on second surface. Find angle of prism :

• A. 60\(^\circ\)
• B. 50\(^\circ\)
• C. 90\(^\circ\)
• D. 30\(^\circ\)

Hint:

In prism problems, the two key relations are Snell's law at both surfaces and the geometric relation \(A = r_1 + r_2\). For special cases like minimum deviation (\(r_1=r_2=A/2\)) or grazing emergence (\(r_2=C\)), these relations simplify the problem significantly.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
A light ray passes through a prism of refractive index \(n\) under the condition of minimum deviation. A thin film of refractive index \(n/2\) is placed on the second face of the prism. Total Internal Reflection (TIR) occurs at the interface between the prism and this film. We need to find the angle of the prism, A.
Step 2: Key Formula or Approach:
1. Condition for minimum deviation: The angle of incidence \(i\) equals the angle of emergence \(e\), and the angles of refraction inside the prism are equal, \(r_1 = r_2\). Also, \(r_1 + r_2 = A\), which simplifies to \(2r = A\) or \(r = A/2\).
2. Condition for Total Internal Reflection (TIR): For a ray traveling from a denser medium (index \(n_d\)) to a rarer medium (index \(n_r\)), TIR occurs if the angle of incidence is greater than or equal to the critical angle \(C\), where \(\sin C = n_r / n_d\).
Step 3: Detailed Explanation:
The light ray is inside the prism (refractive index \(n\)) and is incident on the second face, which is coated with a film of refractive index \(n/2\). Since \(n>n/2\), the prism is the denser medium and the film is the rarer medium.
Let the angle of incidence on the second face be \(r_2\). For TIR to occur at this prism-film interface, the angle \(r_2\) must be greater than or equal to the critical angle \(C\).
The critical angle C for this interface is given by:
\[ \sin C = \frac{\text{Refractive index of rarer medium}}{\text{Refractive index of denser medium}} = \frac{n/2}{n} = \frac{1}{2} \] This gives the critical angle:
\[ C = \sin^{-1}(1/2) = 30^\circ \] So, the condition for TIR is \(r_2 \ge C\), or \(r_2 \ge 30^\circ\).
Now, we use the condition of minimum deviation. At minimum deviation, we have \(r_1 = r_2 = r\).
Also, the prism angle A is related to these angles by \(A = r_1 + r_2\).
Therefore, at minimum deviation, \(A = 2r_2\), or \(r_2 = A/2\).
Combining the two conditions:
\[ \frac{A}{2} \ge 30^\circ \] \[ A \ge 60^\circ \] The question asks for the angle of the prism. The options are discrete values. This implies we are looking for the limiting condition, i.e., the minimum angle A for which TIR is possible. This occurs when \(r_2\) is exactly equal to the critical angle.
\[ A_{min} = 60^\circ \] Therefore, the angle of the prism must be 60\(^\circ\).
Step 4: Final Answer:
The angle of prism is 60\(^\circ\).