Case: Read the passage and answer the question.
Total internal reflection is the reflection of light back into the denser medium at its interface with a rarer medium. For it to occur, light must travel from the denser medium towards the rarer medium, and the angle of incidence in the denser medium must be greater than the critical angle for the pair of media. The critical angle depends on the nature of the media and on the wavelength of light. It can be shown that \( \mu=1/\sin C \), where C is the critical angle.
Total internal reflection is possible only when the ratio of the refractive index of the denser medium to that of the rarer medium is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → more than one
The passage says that for total internal reflection the light must go from the denser medium towards the rarer medium. It also gives \( \mu=1/\sin C \). Here \( \mu \) is the refractive index of the denser medium relative to the rarer one. Since \( \sin C \) can be at most 1, a critical angle exists only when \( \mu\gt1 \).
Option (1) zero — Incorrect. A refractive index ratio can never be zero. The value of \( 1/\sin C \) is always at least 1.
Option (2) one — Incorrect. Equal refractive indices mean the light passes into the second medium without bending. Also, \( \sin C=1 \) gives \( C=90^\circ \), and no angle of incidence can be larger than this.
Option (3) less than one — Incorrect. The light would then go from a rarer into a denser medium. There it bends towards the normal and always enters. Also, \( \sin C=1/\mu \) would be more than 1, so no critical angle exists.
Option (4) more than one — Correct. With \( \mu\gt1 \), \( \sin C=1/\mu \) is less than 1, so a critical angle exists. Light falling at a larger angle in the denser medium is totally reflected.