Unpolarised light is incident on the boundary between two dielectric media, whose dielectric constants are 2.8 (medium -1 ) and 6.8 (medium -2 ), respectively. To satisfy the condition, so that the reflected and refracted rays are perpendicular to each other, the angle of incidence should be \tan ^ - 1( 1 + 10 \over \theta )^1 \over 2 the value of \theta is \underline\hspace2cm . (Given for dielectric media, \mu_r=1 )

Wave Optics PYQ — JEE Main 2023 · 29 Jan · Shift 2 · Q68 with Solution

View written solutionFree

Correct answer: 7

For dielectric media with $\mu_r = 1$ , the refractive index is $n = \sqrt{\varepsilon_r}.$

So, $n_1 = \sqrt{2.8}, \qquad n_2 = \sqrt{6.8}.$

The condition that the reflected and refracted rays are perpendicular is the Brewster condition. Hence, $i_B + r_B = 90^\circ$ and therefore, $\tan i_B = \frac{n_2}{n_1}.$
Substitute the refractive indices: $\tan i_B = \frac{\sqrt{6.8}}{\sqrt{2.8}} = \sqrt{\frac{6.8}{2.8}}.$

Now, $\frac{6.8}{2.8} = \frac{68}{28} = \frac{17}{7}.$ So, $\tan i_B = \sqrt{\frac{17}{7}}.$

The angle is given in the form $\tan^{-1}\left(1+\frac{10}{\theta}\right)^{1/2}.$ Thus, $\sqrt{1+\frac{10}{\theta}} = \sqrt{\frac{17}{7}}.$

Squaring both sides, $1+\frac{10}{\theta} = \frac{17}{7}.$

Solve for $\theta$ : $\frac{10}{\theta} = \frac{17}{7}-1 = \frac{10}{7}.$ Hence, $\frac{10}{\theta} = \frac{10}{7} \implies \theta = 7.$

Therefore, the required integer value is $\boxed{7}.$