If the value of the refractive index of a denser medium with respect to the rarer medium is 1.414, then the critical angle is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 45^\circ \)
At the critical angle \( C \) for light going from the denser to the rarer medium, the angle of refraction is \( 90^\circ \), so Snell's law gives \( \sin C = 1/n \), where \( n \) is the refractive index of the denser medium with respect to the rarer one.
\( \displaystyle \sin C = \frac{1}{1.414} = 0.707 = \frac{1}{\sqrt{2}} \)
\( \displaystyle C = 45^\circ \)