JEE Physics 2023 Question with Solution

Given: dielectric constants are $\varepsilon_1 = 2.8$ and $\varepsilon_2 = 6.8$.

Find: the value of $\theta$ for which the reflected and refracted rays are perpendicular.

Using Brewster's law, the angle of incidence satisfies

$$\tan \theta = \sqrt{\frac{\mu_2}{\mu_1}}$$

where $\mu_1 = \sqrt{\varepsilon_1}$ and $\mu_2 = \sqrt{\varepsilon_2}$.

Given $\varepsilon_1 = 2.8$ and $\varepsilon_2 = 6.8$,

$$\tan \theta = \sqrt{\frac{6.8}{2.8}}$$

Simplifying,

$$\tan \theta = \sqrt{2.43} \approx 1.56$$

Taking the arctangent,

$$\theta = \tan^{-1}(1.56) \approx 7^\circ$$

Therefore, the value of $\theta$ is $7$.

Given: the two media have dielectric constants $2.8$ and $6.8$.

Find: the Brewster angle $\theta$.

The required condition is that the reflected ray and refracted ray are at right angles to each other. This is the Brewster condition. For this case,

$$\tan \theta = \sqrt{\frac{\mu_2}{\mu_1}}$$

Now substitute the given values as used in the solution:

$$\tan \theta = \sqrt{\frac{6.8}{2.8}} = \sqrt{2.43}$$

Hence,

$$\tan \theta \approx 1.56$$

So,

$$\theta = \tan^{-1}(1.56) \approx 7^\circ$$

Thus, the numerical answer is $7$.