Question

A ray of light passes from a medium of refractive index $\sqrt{2}$ to the medium of refractive index $1$. For the ray to just graze along the surface of separation, the angle of incidence should be:

• A. $\dfrac{\pi}{3}$
• B. $\dfrac{\pi}{4}$
• C. $\dfrac{\pi}{6}$
• D. $\dfrac{2\pi}{3}$

Hint:

Grazing emergence means the angle of incidence equals the critical angle, with sin of it equal to n2/n1 = 1/root 2.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
When light goes from a denser medium to a rarer one, it bends away from the normal. The refracted ray grazes the surface (angle of refraction \(90^{\circ}\)) when the angle of incidence equals the critical angle \(\theta_c\).

Step 2: Key Formula or Approach:
Snell's law: \(n_1 \sin\theta_1 = n_2 \sin\theta_2\). For grazing emergence \(\theta_2 = 90^{\circ}\), so \(\sin\theta_c = n_2/n_1\).

Step 3: Detailed Explanation:
Here \(n_1 = \sqrt{2}\) and \(n_2 = 1\):
\[ \sin\theta_c = \frac{1}{\sqrt{2}} \Rightarrow \theta_c = 45^{\circ} = \frac{\pi}{4} \]

Step 4: Why the other options are wrong:
(A) \(\pi/3 = 60^{\circ}\) gives \(\sin 60^{\circ} = 0.866\), which is larger than \(1/\sqrt{2}\). The ray would be totally reflected instead of grazing.
(C) \(\pi/6 = 30^{\circ}\) gives \(\sin 30^{\circ} = 0.5\). The ray would emerge at a smaller angle of refraction, not grazing.
(D) \(2\pi/3 = 120^{\circ}\) is more than \(90^{\circ}\), which is not a valid angle of incidence.

Final Answer:
The angle of incidence is the critical angle \(\pi/4\). This is option (B). \[ \boxed{\theta_c = \frac{\pi}{4}} \]