Calculate the critical angle for total internal reflection when light travels from water ( $n = 1.33$ ) to air ( $n = 1.00$ ).

easy 1 mark Class 10 · Science · Light - Reflection and Refraction

Answer

^{\circ}

The correct answer is 49 $^{\circ}$ — Rounded to the nearest degree, the critical angle is $4 9^{\circ}$ .

Step-by-step solution

The critical angle $c$ is given by $sin c = \frac{n _{2}}{n _{1}}$ , where $n_{1}$ is the refractive index of the denser medium (water) and $n_{2}$ that of the rarer medium (air).
Substitute the values: $sin c = \frac{1.00}{1.33} = 0.7519$ .
Take the inverse sine: $c = sin^{- 1} (0.7519) \approx 48. 8^{\circ}$ .
Rounded to the nearest degree, the critical angle is $4 9^{\circ}$ .

Common mistake: Using

n_{1} / n_{2}

instead of

n_{2} / n_{1}

, which yields a value greater than 1 and is invalid.

Concept tested: Critical angle and total internal reflection