Question

A prism with angle of prism 75° and having refractive index \(\sqrt{3}\) has a slab of refractive index 1.5 kept on one side of the prism as shown. Find angle of incidence such that TIR occurs at slab prism interface. (Given sin 15° = 0.25 and sin 25° = 0.43) :

• A. \(10^\circ<i<20^\circ\)
• B. \(0^\circ<i<25^\circ\)
• C. \(0^\circ<i<15^\circ\)
• D. \(15^\circ<i<25^\circ\)

Hint:

In multi-step optics problems, it's often helpful to work backward from the final condition.
Start with the TIR condition to find the constraint on the internal angle (\(r_2\)), then use the prism geometry to find the constraint on the other internal angle (\(r_1\)), and finally use Snell's law to find the required range for the initial angle of incidence (\(i\)).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the range of the angle of incidence \(i\) on the first face of the prism such that light undergoes Total Internal Reflection (TIR) at the second face, which is in contact with a glass slab.
Step 2: Key Formula or Approach:
1. Condition for TIR: TIR occurs when light travels from a denser medium (\(n_1 = \sqrt{3}\)) to a rarer medium (\(n_2 = 1.5\)) and the angle of incidence at the interface (\(r_2\)) is greater than the critical angle \(C\). The critical angle is given by \(\sin C = \frac{n_2}{n_1}\).
2. Prism Angle Relation: For a prism, the angle of prism \(A\) is related to the angle of refraction at the first face (\(r_1\)) and the angle of incidence at the second face (\(r_2\)) by \(A = r_1 + r_2\).
3. Snell's Law: At the first face, \(n_{air} \sin i = n_1 \sin r_1\).
Step 3: Detailed Explanation:
Part A: Calculate the Critical Angle (C)
Light travels from the prism (\(n_1 = \sqrt{3} \approx 1.732\)) to the slab (\(n_2 = 1.5\)).
\[ \sin C = \frac{n_2}{n_1} = \frac{1.5}{\sqrt{3}} = \frac{3/2}{\sqrt{3}} = \frac{\sqrt{3}}{2} \] \[ C = 60^\circ \] The condition for TIR to occur is \(r_2>C\), so \(r_2>60^\circ\).
Part B: Find the corresponding angle \(r_1\)
Using the prism angle relation \(A = r_1 + r_2\), with \(A = 75^\circ\):
\[ 75^\circ = r_1 + r_2 \] Since we need \(r_2>60^\circ\), we can find the corresponding range for \(r_1\):
\[ r_1 = 75^\circ - r_2<75^\circ - 60^\circ \] \[ r_1<15^\circ \] Part C: Find the corresponding angle of incidence \(i\)
Apply Snell's law at the first face (air to prism):
\[ 1 \cdot \sin i = n_1 \sin r_1 = \sqrt{3} \sin r_1 \] Since we need \(r_1<15^\circ\) (and \(r_1>0\)), we have:
\[ \sin i<\sqrt{3} \sin 15^\circ \] Using the given value \(\sin 15^\circ = 0.25\):
\[ \sin i<\sqrt{3} \times 0.25 \approx 1.732 \times 0.25 \approx 0.433 \] We are also given \(\sin 25^\circ = 0.43\). So, the condition becomes:
\[ \sin i<0.433 \approx \sin 25^\circ \] \[ i<25^\circ \] The angle of incidence must also be positive, so \(i>0\).
Step 4: Final Answer:
The range of the angle of incidence for TIR to occur is \(0^\circ<i<25^\circ\).