Verified Solution • 2025–26 Session moderate

QJEEM19Phcf6e: JEE Main 2019 Physics - Class 12 JEEM19Phcf6e Solution: An electromagnetic wave of intensity 50 Wm –2 enters in a…

An electromagnetic wave of intensity 50 Wm–2 enters in a medium of refractive index 'n' without any loss. The ratio of the magnitudes of electric, and the ratio of the magnitudes of magnetic fields of the wave before and after entering into the medium are respectively, given by:

Multiple Choice Options

  • A (1n,1n)\left( {{1 \over {\sqrt n }},{1 \over {\sqrt n }}} \right)
  • B (n,n)\left( {\sqrt n ,\sqrt n } \right)
  • C (n,1n)\left( {\sqrt n ,{1 \over {\sqrt n }}} \right)
  • D (1n,n)\left( {{1 \over {\sqrt n }},\sqrt n } \right)

Step-by-Step Verified Solution

C =1μ0∈0= {1 \over {\sqrt {{\mu _0}{ \in _0}} }}

V = =1k∈0μ0= {1 \over {\sqrt {k{ \in _0}{\mu _0}} }} [For transparent medium μ\mu r ≈\approx μ\mu 0]

∴\therefore    CV{C \over V} == k=\sqrt k = n

12∈0 E02{1 \over 2} \in {}_0\,E_0^2 C == intensity == 12{1 \over 2} ∈\in 0 kE2v

∴\therefore    E 02_0^2 C == kE2v

⇒\Rightarrow    E02E2=kVC=n2n⇒E0E=n{{E_0^2} \over {{E^2}}} = {{kV} \over C} = {{{n^2}} \over n} \Rightarrow {{{E_0}} \over E} = \sqrt n

similarly

B02C2μ0=B2v2μ0⇒B0B=1n{{B_0^2C} \over {2{\mu _0}}} = {{{B^2}v} \over {2{\mu _0}}} \Rightarrow {{{B_0}} \over B} = {1 \over {\sqrt n }}

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