Critical angle for glass-air interface (n_glass = 1.5):
Asin⁻¹(1/1.5) ≈ 41.8°
BAbout 30°
C60°
D90°
Answer & Solution
Correct answer: A. sin⁻¹(1/1.5) ≈ 41.8°
sin θ_c = 1/n = 1/1.5 = 0.667. θ_c ≈ 41.8°.
Used in 45-45-90 prisms (binoculars, periscopes): TIR replaces silvered mirrors.
Keep practising Ray and Wave Optics
Browse all Ray and Wave Optics questions for JEE Advanced, or explore more Physics topics.
Related questions
True or false: opticsReflection or refraction?For thin lens, when image distance equals object distance (|v| = |u|), object position is:Diffraction effects are more noticeable when the obstacle/aperture size is:Speed of light in glass (n = 1.5) is approximately:Refractive index of medium 1 with respect to medium 2 (denoted ₁n₂ or n₁/n₂). If ₁n₂ = 1.5For thin-film interference (oil on water), constructive interference for reflection conditTwo coherent sources of intensities I₁ and I₂ interfere. Maximum intensity in pattern: