JEE Main 2021 — Electromagnetic Waves Question with Solution

Given, electric field,

E = 20 cos(2 $$\times$$ 10¹⁰t $$-$$ 200 x) V/m

Comparing with the standard equation,

E = E₀ cos($$\omega$$t $$-$$ kx) V/m, we get

Wave constant, k = 200

Angular frequency, $$\omega$$ = 2 $$\times$$ 10¹⁰ rad/s

Speed of the wave, $$v=\frac{\omega }{k}=\frac{2\times {10}^{10}}{200}=10^8$$ m/s

Refractive index, $$\mu =\frac{c}{v}=\frac{3\times {10}^8}{{10}^8}=3$$

As we know the relation between the refractive index and dielectric constant,

$$\mu =\sqrt{{\epsilon }_r{\mu }_r}$$

Substituting the value in the above equations, we get

$$3=\sqrt{{\epsilon }_r(1)}$$

$${\epsilon }_r=9$$

Thus, the dielectric constant of the medium is 9.