A light ray is incident normal to one face of a right-angled isosceles prism and is totally internally reflected at the glass air interface. If the angle of reflection is \( \pi/4 \), what can be concluded about the refractive index?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \mu \gt \sqrt{2} \)
The ray enters the first face without bending and meets the hypotenuse at \( 45^\circ \), which is \( \pi/4 \). For total internal reflection this angle of incidence must exceed the critical angle \( C \) of the glass.
\( \displaystyle \sin C = \frac{1}{\mu} \quad\text{and}\quad 45^\circ \gt C \;\Rightarrow\; \sin 45^\circ \gt \frac{1}{\mu} \)
\( \displaystyle \mu \gt \frac{1}{\sin 45^\circ} = \sqrt{2} \)
At exactly \( \mu = \sqrt{2} \) the angle of incidence equals the critical angle and the refracted ray just grazes the surface, so reflection is not yet total.
Note: This question is from NCERT (2026-27). NTA sets numerical questions from NCERT too, so do not ignore NCERT while practising numericals.
NCERT (2026-27):
Total Internal Reflection in Nature and Its Technological Applications – “… Prisms designed to bend light by 90° or by 180° make use of total internal reflection [Fig. 9.13(a) and (b)]. … In the first two cases, the critical angle ic for the material of the prism must be less than 45°.”