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QJEEAdv22Ph0211: JEE Advanced 2022 Physics - Class 12 JEEAdv22Ph0211 Solution: Consider a configuration of identical units, each consisting of …

Consider a configuration of nn identical units, each consisting of three layers. The first layer is a column of air of height h=13 cmh=\frac{1}{3} \mathrm{~cm} , and the second and third layers are of equal thickness d=d= 3−12 cm\frac{\sqrt{3}-1}{2} \mathrm{~cm} , and refractive indices μ1=32\mu_{1}=\sqrt{\frac{3}{2}} and μ2=3\mu_{2}=\sqrt{3} , respectively. A light source 0 is placed on the top of the first unit, as shown in the figure. A ray of light from 0 is incident on the second layer of the first unit at an angle of θ=60∘\theta=60^{\circ} to the normal. For a specific value of nn , the ray of light emerges from the bottom of the configuration at a distance l=83 cml=\frac{8}{\sqrt{3}} \mathrm{~cm} , as shown in the figure. The value of nn is ________.

JEE Advanced 2022 Paper 2 Online Physics - Geometrical Optics Question 16 English

Step-by-Step Verified Solution

JEE Advanced 2022 Paper 2 Online Physics - Geometrical Optics Question 16 English Explanation
amp;x1=13×tan⁡60amp;=13 cm\begin{aligned}&x_1 =\frac{1}{3} \times \tan 60 \\\\& =\frac{1}{\sqrt{3}} \mathrm{~cm}\end{aligned}

By Snell's law

amp;1×sin⁡60=32sin⁡θ2amp;⇒θ2=45oamp;∴x2=d\begin{aligned}&1 \times \sin 60 =\frac{\sqrt{3}}{\sqrt{2}} \sin \theta_2 \\\\& \Rightarrow \theta_2 =45^ \mathrm{o} \\\\&\therefore x_2 =d\end{aligned}

Again, by applying Snell's law

amp;32×12=3×sin⁡θ3amp;⇒θ2=30∘amp;∴x3=d3\begin{aligned}&\sqrt{\frac{3}{2} \times \frac{1}{\sqrt{2}}} =\sqrt{3} \times \sin \theta_3 \\\\&\Rightarrow \theta_2 =30^{\circ} \\\\& \therefore x_3 =\frac{d}{\sqrt{3}}\end{aligned}

Also
amp;x1+x2+x3=13+d+d3amp;=13+(3−12)+3−123amp;=2+(3−3)+3−123amp;=423amp;x1+x2+x3=23 cmamp;And,  η=lx1+x2+x3amp;η=8/32/3=4amp;η=4\begin{aligned}& x_1+x_2+x_3 =\frac{1}{\sqrt{3}}+d+\frac{d}{\sqrt{3}} \\\\& =\frac{1}{\sqrt{3}}+\left(\frac{\sqrt{3}-1}{2}\right)+\frac{\sqrt{3}-1}{2 \sqrt{3}} \\\\& =\frac{2+(3-\sqrt{3})+\sqrt{3}-1}{2 \sqrt{3}} \\\\& =\frac{4}{2 \sqrt{3}} \\\\& x_1+x_2+x_3 =\frac{2}{\sqrt{3}} \mathrm{~cm} \\\\& \text {And,}~~ \eta =\frac{l}{x_1+x_2+x_3} \\\\&\eta =\frac{8 / \sqrt{3}}{2 / \sqrt{3}}=4 \\\\&\eta =4\end{aligned}

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