n= refractive index =ϵrμr
I=21ϵ0E02C
I′=21ϵE2V
I′=0.96I
21ϵE2v=0.9621ϵ0E02C
E=0.96ϵ0ϵrϵ0vcE0
(For most of the transparent medium μr≈1 )
⇒E=0.96.ϵrμr1.vcE0
=0.96.n1.nE0
E=n0.96E0E=1.50.96×30=24mV
JEE Main 2019 — Physics Optics
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 30mV, then the amplitude of the electric field for the wave propagating in the glass medium will be:
Held on 12 Jan 2019 · Verified 6 Jul 2026.
30mV
6mV
24mV
10mV
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