Question

A glass slab has refractive index $\mu$ with respect to air and the critical angle for a ray of light in going from glass to air is $\theta$. If a ray of light is incident from air on the glass with angle of incidence $\theta$, then the corresponding angle of refraction is

• A. $\sin^{-1}\left(\frac{1}{\sqrt{\mu}}\right)$
• B. $\sin^{-1}\left(\frac{1}{\mu}\right)$
• C. $\sin^{-1}\left(\frac{1}{\mu^2}\right)$
• D. $90^\circ$

Hint:

Whenever a problem sets the launch angle equal to the critical angle $\theta$, you are essentially multiplying by a factor of $\frac{1}{\mu}$ twice—once to define the angle ($\sin\theta = \frac{1}{\mu}$) and once when applying Snell's law ($\sin r = \frac{\sin\theta}{\mu}$). This naturally leads to an effective $\frac{1}{\mu^2}$ argument.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem involves two scenarios. First, light moves from glass to air, defining the critical angle $\theta$.
Second, a different light ray travels in the opposite direction (air to glass), hitting the interface with an angle of incidence equal to that same angle $\theta$. We must calculate its resulting angle of refraction $r$.

Step 2: Key Formula or Approach:
1. The critical angle equation when exiting a medium of refractive index $\mu$ into air ($\mu_{air} = 1$) is:
$$\sin\theta = \frac{1}{\mu}$$ 2. Snell's law for light entering glass from air is:
$$\frac{\sin i}{\sin r} = \mu \implies \sin r = \frac{\sin i}{\mu}$$

Step 3: Detailed Explanation:
From the definition of the critical angle given in the problem, we establish:
$$\sin\theta = \frac{1}{\mu}$$ Now consider the reverse path where the ray hits the glass surface from air.
The problem specifies the angle of incidence is equal to the critical angle, so $i = \theta$.
Substitute $i = \theta$ into Snell's law:
$$\sin r = \frac{\sin\theta}{\mu}$$ Substitute our initial value of $\sin\theta = \frac{1}{\mu}$ directly into this expression:
$$\sin r = \frac{\left(\frac{1}{\mu}\right)}{\mu} = \frac{1}{\mu^2}$$ To solve for the final angle of refraction $r$, take the inverse sine of both sides:
$$r = \sin^{-1}\left(\frac{1}{\mu^2}\right)$$

Step 4: Final Answer:
The corresponding angle of refraction is $\sin^{-1}\left(\frac{1}{\mu^2}\right)$, which matches option (C).