Ray Optics

JEE Main level, single correct, medium

111 of 295

Question 111 of 295

A light wave is incident normally on a glass slab of refractive index 1.5.1.5. If 44% of light gets reflected and the amplitude of the electric field of the incident light is 30Vm,30\dfrac{V}{m}, then the amplitude of the electric field for the wave propagating in the glass medium will be:

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Answer: (C) 24Vm24\dfrac{V}{m}

n=n= refractive index =ϵrμr=\sqrt{{\epsilon }_{r}{\mu }_{r}}
I=12ϵ0E02CI=\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}C

I′=12ϵE2V{I}^{'}=\dfrac{1}{2}\epsilon {E}^{2}V

I′=0.96I{I}^{'}=0.96I

12ϵE2v=0.9612ϵ0E02C\dfrac{1}{2}\epsilon {E}^{2}v=0.96\dfrac{1}{2}{\epsilon }_{0}{E}_{0}^{2}C

E=0.96ϵ0ϵ0ϵrcvE0E=\sqrt{0.96}\sqrt{\dfrac{{\epsilon }_{0}}{{\epsilon }_{0}{\epsilon }_{r}}}\sqrt{\dfrac{c}{v}}{E}_{0}

(For most of the transparent medium μr≈1{\mu }_{r}\approx 1 )
⇒E=0.96.1ϵrμr.cvE0\Rightarrow E=\sqrt{0.96}.\sqrt{\dfrac{1}{{\epsilon }_{r}{\mu }_{r}}}.\sqrt{\dfrac{c}{v}}{E}_{0}

=0.96.1n.nE0=\sqrt{0.96}.\dfrac{1}{n}.\sqrt{n}{E}_{0}

E=0.96nE0E=0.961.5×30=24VmE=\sqrt{\dfrac{0.96}{n}}{E}_{0} E=\sqrt{\dfrac{0.96}{1.5}}\times 30=24\dfrac{V}{m}

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