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: Tie both events to one constant, $\mu$.
The same refractive index $\mu$ governs both the critical-angle condition and the new air-to-glass refraction, so we will express both through $\mu$ and link them.
Step 2: Critical angle gives $\sin\theta$.
For glass-to-air, total internal reflection begins when $\sin\theta = \dfrac{1}{\mu}$. Keep this handy.
Step 3: Snell's law for the new ray.
Air-to-glass with incidence angle equal to $\theta$: $\sin\theta = \mu \sin r$, where $r$ is the refraction angle inside the glass.
Step 4: Solve for $\sin r$.
$\sin r = \dfrac{\sin\theta}{\mu}$.
Step 5: Substitute the critical-angle result.
Replacing $\sin\theta$ with $\dfrac{1}{\mu}$ gives $\sin r = \dfrac{1/\mu}{\mu} = \dfrac{1}{\mu^{2}}$.
Step 6: Conclude.
Therefore the refraction angle is \[ \boxed{r = \sin^{-1}\!\left(\frac{1}{\mu^{2}}\right)} \]