Wave Optics

JEE Main level, numerical answer, medium

171 of 191

Question 171 of 191

Unpolarised light is incident on the boundary between two dielectric media, whose dielectric constants are 2.82.8 (medium –11) and 6.86.8 (medium – 22), 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+10θ)12{\tan }^{-1}{\left(1+\dfrac{10}{\theta }\right)}^{\dfrac{1}{2}}, the value of θ\theta is ______.
(Given for dielectric media, μr=1{\mu }_{r}=1)

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Answer: 7

As given in the question, we can write
μ1=ϵrμr=2.8×1=2.8{\mu }_{1}=\sqrt{{\epsilon }_{r}{\mu }_{r}}=\sqrt{2.8\times 1}=\sqrt{2.8} and
μ2=ϵrμr=6.8×1=6.8{\mu }_{2}=\sqrt{{\epsilon }_{r}{\mu }_{r}}=\sqrt{6.8\times 1}=\sqrt{6.8}
Now applying Snell's law and considering r=90∘−ir=90^{\circ}-i, we get
μ1sin⁡i=μ2cos⁡i{\mu }_{1}\sin i={\mu }_{2}\cos i
⇒tan⁡i=μ2μ1=6.82.8\Rightarrow \tan i=\dfrac{{\mu }_{2}}{{\mu }_{1}}=\sqrt{\dfrac{6.8}{2.8}}
⇒tan⁡i=(2.8+42.8)12\Rightarrow \tan i={\left(\dfrac{2.8+4}{2.8}\right)}^{\dfrac{1}{2}}
⇒i=tan⁡−1(1+107)12\Rightarrow i={\tan }^{-1}{\left(1+\dfrac{10}{7}\right)}^{\dfrac{1}{2}}
Therefore, θ=7\theta =7.

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