Electromagnetic Waves

JEE Main level, single correct, medium

11 of 165

Question 11 of 165

The electric field of an electromagnetic wave travelling through a medium is given by
E⃗(x,t)=25sin⁡(2.0×1015t−107x)n^\vec{E}(x, t)=25 \sin \left(2.0 \times 10^{15} t-10^{7} x\right) \hat{n}
then the refractive index of the medium is ____\_\_\_\_.
(All given measurement are in SI units)

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Answer: (A) 1.5

The electric field is E(x,t)=25sin⁡(2.0×1015t−107x)E(x,t) = 25\sin(2.0 \times 10^{15}t - 10^7 x) V/m.
This is in the standard form E=E0sin⁡(ωt−kx)E = E_0\sin(\omega t - kx) where angular frequency ω=2.0×1015\omega = 2.0 \times 10^{15} rad/s and wave number k=107k = 10^7 rad/m.
The speed of the wave in the medium is v=ωk=2.0×1015107=2.0×108v = \frac{\omega}{k} = \frac{2.0 \times 10^{15}}{10^7} = 2.0 \times 10^8 m/s.
The refractive index is defined as n=cvn = \frac{c}{v} where c=3.0×108c = 3.0 \times 10^8 m/s is the speed of light in vacuum.
Therefore, n=3.0×1082.0×108=1.5n = \frac{3.0 \times 10^8}{2.0 \times 10^8} = 1.5

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