JEE Main 12 Jan 2019 Morning Question 15
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 $ 30V,{m^{-1}}, $ then the amplitude of the electric field for the wave propagating in the glass medium will be [JEE Main Online Paper Held On 12-Jan-2019 Morning]
Options:
A) $ 24V,{m^{-1}} $
B) $ 10V,{m^{-1}} $
C) $ 30V,{m^{-1}} $
D) $ 6V,{m^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ E _{ot} $ and $ E _{oi} $ be the perpendicular components of the electric field for transmitted and incident waves respectively.
$ \frac{E _{ot}}{E _{oi}}=\frac{2{\mu_1}}{{\mu_1}+{\mu_2}}=\frac{2(1)}{1+1.5}=\frac{4}{5} $
$ \Rightarrow $ $ E _{ot}=30\times \frac{4}{5}=24V,{m^{-1}} $
