Question

If the critical angle in a medium is \( \alpha \), then for total internal reflection the angle of incidence \( \beta \) should be:

• A. \( \beta < \alpha \)
• B. \( \beta > \alpha \)
• C. \( \beta = \alpha \)
• D. \( \beta \leq \alpha \)

Hint:

Total internal reflection happens only when the ray goes from denser to rarer medium and the angle of incidence exceeds the critical angle.

Answer 1

The Correct Option is B

Step 1: Start from Snell's law at the denser-to-rarer boundary.
For a ray going from a medium of refractive index \(n\) (denser) into air, \(n\sin\beta = \sin r\), where \(\beta\) is the incidence angle and \(r\) the refraction angle. Since \(n>1\), \(r\) is always larger than \(\beta\).

Step 2: Push the refracted ray to its limit.
As \(\beta\) increases, \(r\) increases faster and reaches its maximum possible value \(90^\circ\). The incidence angle at which this happens is defined as the critical angle \(\alpha\), so \(n\sin\alpha = \sin 90^\circ = 1\).

Step 3: Go beyond the limit.
If we make \(\beta\) even larger than \(\alpha\), the equation \(\sin r = n\sin\beta\) would demand \(\sin r > 1\), which is mathematically impossible. Nature responds by not transmitting the ray at all; instead it reflects the entire beam back inside the denser medium.

Step 4: Conclusion.
This complete reflection (total internal reflection) begins only once \(\beta\) exceeds \(\alpha\). Hence the required condition is that the incidence angle be greater than the critical angle.

\[\boxed{\beta > \alpha}\]