JEE Physics 2026 Question with Solution

Given: Refractive index of incident medium is $n_1 = 2$ and refractive index of refracted medium is $n_2 = 2\sqrt{3}$.

Find: The angle of incidence for which the reflected and refracted rays are perpendicular.

If the reflected and refracted rays are perpendicular, then the angle of incidence is the Brewster angle.

Using Brewster’s law,

$$\tan i_B = \frac{n_2}{n_1}$$

Substituting the given refractive indices,

$$\tan i_B = \frac{2\sqrt{3}}{2} = \sqrt{3}$$

Therefore,

$$i_B = 60^\circ$$

Hence, the angle of incidence is $60^\circ$ and the correct option is D.

Given: The two dielectric media have refractive indices $2$ and $2\sqrt{3}$.

Find: The required angle of incidence.

Step 1: Using the condition for perpendicular reflected and refracted rays.

When reflected and refracted rays are perpendicular to each other, the angle of incidence equals the Brewster angle.

Step 2: Applying Brewster’s law.

$$\tan i_B = \dfrac{n_2}{n_1}$$

Step 3: Substituting the given refractive indices.

$$\tan i_B = \dfrac{2\sqrt{3}}{2} = \sqrt{3}$$

Step 4: Determining the angle of incidence.

$$i_B = 60^\circ$$

Step 5: Final conclusion. The angle of incidence is $60^\circ$, corresponding to option D.