JEE Physics 2026 Question with Solution

Given:

• Electric field of the plane electromagnetic wave:

$$E = E_0 \sin (kx - \omega t)$$

• From the given equation,

$$k = 3 \times 10^6 \, \text{m}^{-1}$$ $$\omega = 4.5 \times 10^{14} \, \text{rad/s}$$

Find: The dielectric constant of the non-magnetic medium.

The general equation of a plane electromagnetic wave is:

$$E = E_0 \sin (kx - \omega t)$$

Comparing with the given equation gives the values of $k$ and $\omega$ above.

Step 1: Calculate the speed of the wave in the medium.

$$v = \frac{\omega}{k} = \frac{4.5 \times 10^{14}}{3 \times 10^6}$$ $$v = 1.5 \times 10^8 \, \text{m/s}$$

Step 2: Find refractive index of the medium. Speed of light in vacuum:

$$c = 3 \times 10^8 \, \text{m/s}$$ $$n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$$

Step 3: Calculate dielectric constant. Since the medium is non-magnetic:

$$\varepsilon_r = n^2 = 2^2 = 4$$

Therefore, the dielectric constant of the medium is $4$.

Given: For a non-magnetic medium, dielectric constant is equal to the square of the refractive index.

Find: The value of dielectric constant.

First determine the wave speed from the wave equation coefficients:

$$v = \frac{\omega}{k}$$

Substituting,

$$v = \frac{4.5 \times 10^{14}}{3 \times 10^6} = 1.5 \times 10^8 \, \text{m/s}$$

Now compare with the speed of light in vacuum:

$$n = \frac{c}{v} = \frac{3 \times 10^8}{1.5 \times 10^8} = 2$$

For a non-magnetic medium,

$$\varepsilon_r = n^2$$

Hence,

$$\varepsilon_r = 4$$

So, the required numerical answer is $4$.