Question

A ray is incident from a medium of refractive index \(2\) into a medium of refractive index \(1\). The critical angle is

• A. \(30^\circ\)
• B. \(60^\circ\)
• C. \(45^\circ\)
• D. \(90^\circ\)

Hint:

Critical angle exists only when light travels from a denser medium to a rarer medium. The formula is: \[ \sin C=\frac{n_2}{n_1} \] where \(n_1\gt n_2\).

Answer 1

The Correct Option is A

Step 1: Recognise the situation.
Light goes from a denser medium ($n_1 = 2$) into a rarer medium ($n_2 = 1$). When light moves dense to rare, total internal reflection becomes possible beyond a special angle called the critical angle.
Step 2: Apply Snell's law at the critical angle.
At the critical angle $C$, the refracted ray grazes along the boundary, so the angle of refraction is $90^\circ$: \[ n_1\sin C = n_2\sin 90^\circ. \]
Step 3: Simplify.
Since $\sin 90^\circ = 1$, \[ \sin C = \frac{n_2}{n_1}. \]
Step 4: Substitute the indices.
\[ \sin C = \frac{1}{2}. \]
Step 5: Find the angle.
\[ C = \sin^{-1}\!\left(\frac{1}{2}\right) = 30^\circ. \]
Step 6: Conclude.
Thus the critical angle for this pair of media is $30^\circ$.
\[ \boxed{30^\circ} \]