Total Internal Reflection
Light travelling inside a glass block of refractive index 1.5 strikes the boundary with the surrounding air. Beyond what angle of incidence at this glass-air interface will the light undergo total internal reflection?
Select the correct option:
Solution
41.8°
Total internal reflection occurs when light moving from a denser to a rarer medium meets the boundary at an angle exceeding the critical angle, for which the refracted ray would graze along the surface at 90°. The critical angle satisfies sinθc=n1 when the rarer medium is air, where n is the refractive index of the denser medium. Here n=1.5, so sinθc=1.51=0.667, giving θc=sin−1(0.667)≈41.8°. For any incidence greater than this, no refracted ray exists and the light is wholly reflected back into the glass with negligible loss. This loss-free reflection is precisely what makes total internal reflection superior to ordinary metallic mirrors for guiding light, since metal coatings always absorb a small fraction. The phenomenon requires light to originate in the denser medium and meet the boundary beyond the threshold, conditions met inside optical fibres where repeated total internal reflection confines signals over kilometres. The value 48.6° corresponds mistakenly to n=4/3 (water), not glass. The value 30° is wrong because sin30°=0.5 would imply n=2. The value 60° is wrong as it exceeds the true critical angle and merely names a valid TIR incidence rather than the threshold. This is the NCERT basis for optical fibres and prisms. A check confirms a larger refractive index gives a smaller critical angle, as 1.5 sensibly yields about 42°.
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About This Question
- Subject
- physics
- Chapter
- optics
- Topic
- total internal reflection
- Difficulty
- Medium
- Year
- 2025
Solution
Correct Answer:
41.8°
Total internal reflection occurs when light moving from a denser to a rarer medium meets the boundary at an angle exceeding the critical angle, for which the refracted ray would graze along the surface at 90°. The critical angle satisfies sinθc=n1 when the rarer medium is air, where n is the refractive index of the denser medium. Here n=1.5, so sinθc=1.51=0.667, giving θc=sin−1(0.667)≈41.8°. For any incidence greater than this, no refracted ray exists and the light is wholly reflected back into the glass with negligible loss. This loss-free reflection is precisely what makes total internal reflection superior to ordinary metallic mirrors for guiding light, since metal coatings always absorb a small fraction. The phenomenon requires light to originate in the denser medium and meet the boundary beyond the threshold, conditions met inside optical fibres where repeated total internal reflection confines signals over kilometres. The value 48.6° corresponds mistakenly to n=4/3 (water), not glass. The value 30° is wrong because sin30°=0.5 would imply n=2. The value 60° is wrong as it exceeds the true critical angle and merely names a valid TIR incidence rather than the threshold. This is the NCERT basis for optical fibres and prisms. A check confirms a larger refractive index gives a smaller critical angle, as 1.5 sensibly yields about 42°.
This medium difficulty physics question is from the chapter optics, covering the topic of total internal reflection. It appeared in the 2025 exam.
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