A ray of light ravels from a liquid of refractive index `mu` to air. If the incident beam is rotating at a rate `omega` , what is the angular sped of the refracted beam at the instant the angle of incidence is `30^@` ? (Given ` mu=sqrt (2), omega =1//sqrt(6) rad s^(-1)`).

Explanation

To solve the problem, we need to find the angular speed of the refracted beam (denoted as \( \omega_r \)) when the incident beam is rotating at a rate \( \omega \) and the angle of incidence \( \theta \) is \( 30^\circ \). We are given the refractive index \( \mu = \sqrt{2} \) and \( \omega = \frac{1}{\sqrt{6}} \, \text{rad/s} \).

### Step 1: Use Snell's Law
According to Snell's Law, we have:
\[ \sin r = \mu \cdot \sin \theta \]
where \( r \) is the angle of refraction, \( \mu \) is the refractive index of the liquid, and \( \theta \) is the angle of incidence.
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