JEE Physics 2026 Question with Solution

Given: The electric field is $\vec{E}(x,t)=25\sin(2\times10^{15}t-10^{7}x)\,\hat{n}.$ Find: The refractive index of the medium.

Compare the given wave with the standard form:

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

So,

$$\omega=2\times10^{15} \; \text{rad s}^{-1}, \qquad k=10^{7} \; \text{m}^{-1}$$

The wave speed in the medium is:

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

Substituting the values,

$$v=\frac{2\times10^{15}}{10^{7}}=2\times10^{8} \; \text{m s}^{-1}$$

Now the refractive index is:

$$n=\frac{c}{v}$$

with $c=3\times10^{8} \; \text{m s}^{-1}.$ Therefore,

$$n=\frac{3\times10^{8}}{2\times10^{8}}=1.5$$

Therefore, the refractive index of the medium is $1.5$, so the correct option is B.

Given: The wave equation is $\vec{E}(x,t)=25\sin(2\times10^{15}t-10^{7}x)\,\hat{n}.$ Find: The refractive index.

Read $\omega$ and $k$ directly by comparing with $E_0\sin(\omega t-kx).$ Then use:

$$v=\frac{\omega}{k}, \qquad n=\frac{c}{v}$$

So,

$$v=\frac{2\times10^{15}}{10^{7}}=2\times10^{8} \; \text{m s}^{-1}$$

and hence,

$$n=\frac{3\times10^{8}}{2\times10^{8}}=1.5$$

This works because for a sinusoidal wave, the coefficient of $t$ is the angular frequency and the coefficient of $x$ is the wave number. Thus, the correct option is B.