A light wave is incident normally on a glass slab of refractive index 1.5 . If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V m , then the amplitude

Question

A light wave is incident normally on a glass slab of refractive index 1.5 . If 4 % of light gets reflected and the amplitude of the electric field of the incident light is 30 V m , then the amplitude of the electric field for the wave propagating in the glass medium will be:

TestHub · Accepted Answer

n = refractive index = ε r μ r I = 1 2 ε 0 E 0 2 C I ′ = 1 2 ε E 2 V I ′ = 0.96 I 1 2 ε E 2 v = 0.96 1 2 ε 0 E 0 2 C E = 0.96 ε 0 ε 0 ε r c v E 0 (For most of the transparent medium μ r ≈ 1 ) ⇒ E = 0.96 . 1 ε r μ r . c v E 0 = 0.96 . 1 n . n E 0 E = 0.96 n E 0 E = 0 . 96 1 . 5 × 30 = 24 V m

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