Practice question
Question
What is the refractive index of a medium if the critical angle for light passing into air is \(
48.6^\circ \)?
Explanation
**Frequency of light** does not change on refraction, c = f λ, when enters denser medium speed decreases, wavelength decreases λ' = v/f = c/(n f) = λ/n, frequency same, for λ=480 nm f=c/λ=3×10⁸/480×10⁻⁹=6.25×10¹⁴ Hz, for 630 nm f=4.76×10¹⁴ Hz, for 540 nm f=5.56×10¹⁴ Hz, refractive index determines bending. sin i_c = (n₂/n₁) , where n₂ = 1.0 (air), i_c = 48.6° . sin 48.6° ≈ 0.75 , n₁ = (1.0/0.75) ≈ 1.33 . Using Δ = d sinθ, y = n λ D/d, a sinθ = n λ, I = I₀ cos²θ, n = c/v, sinC = 1/n, λ' = λ/n and A =
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