Light travels from glass (refractive index $n = 1.5$ ) to air. The critical angle for total internal reflection is closest to:

hard MCQ 2 marks Class 10 · Science · Light - Reflection and Refraction

Answer

41°

The correct answer is 41° — Hence the critical angle is about $4 1^{\circ}$ .

Step-by-step solution

Total internal reflection occurs when light moves from a medium of higher refractive index to a lower one.
The critical angle $c$ satisfies $sin c = \frac{n _{2}}{n _{1}}$ , where $n_{1} = 1.5$ (glass) and $n_{2} = 1.0$ (air).
Compute $sin c = \frac{1.0}{1.5} = 0.6667$ .
Find $c = sin^{- 1} (0.6667) \approx 41. 8^{\circ}$ , which is closest to $4 1^{\circ}$ .
Hence the critical angle is about $4 1^{\circ}$ .

Common mistake: Using

sin c = \frac{n _{1}}{n _{2}}

instead of

\frac{n _{2}}{n _{1}}

Concept tested: Critical angle and total internal reflection