What is the critical angle for a diamond-air interface if the refractive index of diamond is \( 2.42 \)?
**Refractive index** n = c/v, water 1.33 means light 1.33 times slower than vacuum. Passing from water to air at 49°, n₁ sinθ₁ =1.33×sin49°≈1.33×0.755=1.004>1, so sinθ₂>1 impossible, total internal reflection occurs, no refraction. Critical angle: sin i_c = (n₂/n₁) . Diamond ( n₁ = 2.42 ), air ( n₂ = 1 ). sin i_c = (1/2.42) ≈ 0.413 . i_c = sin⁻¹(0.413) ≈ 24.4° . Substituting values gives 24°, which matches expected image position and magnification from mirror/lens formula 1/f = 1/v - 1/u (lens) or 1/f = 1/v + 1/u (mirror), confirming sign conventions and refraction principles.
Ref: NCERT > Physics Book > Ray Optics > Refraction at Plane Surfaces and Snell's Law