A ray of light MN is incident normally on the face corresponding with side AB of a prism with an isosceles right-angled triangular base ABC. Trace the path of the ray as it passes through the prism when the refractive index of the prism material is \( \sqrt{2} \), and \( \sqrt{3} \).
Show Hint
To calculate the refraction angles inside a prism, use Snell’s Law at each interface. The refractive index determines how much the ray bends when passing through different materials.
Step 1: Ray at first surface (AB).
The ray \(MN\) is incident normally on face AB.
Angle of incidence \( = 0^\circ \).
Hence, no refraction occurs at AB.
The ray enters the prism undeviated.
Step 2: Ray inside the prism.
Since the prism is an isosceles right-angled prism, angle between faces AB and BC is \(45^\circ\).
Therefore, the ray strikes face BC with angle of incidence:
\[
i = 45^\circ
\]
Step 3: Case 1 — Refractive index \( n = \sqrt{2} \).
Critical angle \(C\) is given by:
\[
\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}
\]
\[
C = 45^\circ
\]
Since angle of incidence \(i = 45^\circ = C\),
the refracted ray emerges along the surface of BC.
Thus, the angle of refraction is:
\[
r = 90^\circ
\]
Step 4: Case 2 — Refractive index \( n = \sqrt{3} \).
Critical angle:
\[
\sin C = \frac{1}{\sqrt{3}}
\]
\[
C \approx 35.3^\circ
\]
Since \( i = 45^\circ > C \),
total internal reflection occurs at face BC.
The ray is reflected inside the prism and will then strike face AC normally,
emerging undeviated from that face.
Final Conclusion:
For \( n = \sqrt{2} \): Ray grazes along face BC.
For \( n = \sqrt{3} \): Total internal reflection occurs at BC and the ray finally emerges from face AC.
