JEE Advance - Physics (2020 - Paper 2 Offline - No. 2)
A large square container with thin transparent vertical walls and filled with water
(refractive index $${4 \over 3}$$) is kept on a horizontal table. A student holds a thin straight wire vertically
inside the water 12 cm from one of its corners, as shown schematically in the figure. Looking at the
wire from this corner, another student sees two images of the wire, located symmetrically on each
side of the line of sight as shown. The separation (in cm) between these images is ;
Answer
4
Explanation
Consider ray of light from object $\mathrm{O}$ moves along $\mathrm{OB}$ at an angle of incidence $i$ and refracted through air parallel to line of sight, that is, $\angle r=45^{\circ}$. Another ray slight above $\mathrm{OA}$ is refracted along $\mathrm{AC}$ and image appears to be formed at $I_1$ having height $h$.
At point B; from Snell's law
$$ \begin{aligned} & \frac{\sin i}{\sin 45}=\frac{1}{4 / 3} \Rightarrow \sin i=\frac{3}{4} \sin 45^{\circ} \\\\ & \Rightarrow \sin =\frac{3}{4} \times \frac{1}{\sqrt{2}}=\frac{3}{4 \times 1.4142}=0.5303 \end{aligned} $$
So, $i=32.03^{\circ}$.
In $\triangle \mathrm{OAB}, \quad \angle B=\left(90^{\circ}+i\right)=\left(90^{\circ}+32.03^{\circ}\right) \Rightarrow 122.03^{\circ}$.
$$ \begin{aligned} & \angle A=45^{\circ} \\\\ & \angle O=\left(180-45^{\circ}-122.03^{\circ}\right) \cong 13^{\circ} \end{aligned} $$
From sine law, we have
$$ \begin{aligned} & \frac{12}{\sin 122^{\circ}}=\frac{\mathrm{AB}}{\sin 13^{\circ}} \Rightarrow \mathrm{AB}=\frac{12}{\sin 122^{\circ}} \times \sin 13^{\circ} \\\\ & \mathrm{AB}=\frac{12}{0.8480} \times 0.2249=3.1833 \mathrm{~cm} \end{aligned} $$
In $\triangle \mathrm{BAE}$,
$$ \begin{aligned} & \frac{\mathrm{BE}}{\mathrm{AB}}=\sin 45^{\circ} \Rightarrow \mathrm{BE}=\mathrm{AB} \sin 45^{\circ} \\\\ & \mathrm{BE}=31833 \times \frac{1}{\sqrt{2}}=\frac{3.1833}{1.4142}=2.25 \mathrm{~cm} \end{aligned} $$
Since, $\mathrm{BE}=\mathrm{FG}$ is position of image $I_1$ at upper half of square, $h=2.25 \mathrm{~cm}$.
Similarly, in lower half of square image $I_2$ will be at a distance of $2.25 \mathrm{~cm}$.
$$ I_1 I_2=2 \times 2.25=4.50 \mathrm{~cm}=4.00 \mathrm{~cm} $$
At point B; from Snell's law
$$ \begin{aligned} & \frac{\sin i}{\sin 45}=\frac{1}{4 / 3} \Rightarrow \sin i=\frac{3}{4} \sin 45^{\circ} \\\\ & \Rightarrow \sin =\frac{3}{4} \times \frac{1}{\sqrt{2}}=\frac{3}{4 \times 1.4142}=0.5303 \end{aligned} $$
So, $i=32.03^{\circ}$.
In $\triangle \mathrm{OAB}, \quad \angle B=\left(90^{\circ}+i\right)=\left(90^{\circ}+32.03^{\circ}\right) \Rightarrow 122.03^{\circ}$.
$$ \begin{aligned} & \angle A=45^{\circ} \\\\ & \angle O=\left(180-45^{\circ}-122.03^{\circ}\right) \cong 13^{\circ} \end{aligned} $$
From sine law, we have
$$ \begin{aligned} & \frac{12}{\sin 122^{\circ}}=\frac{\mathrm{AB}}{\sin 13^{\circ}} \Rightarrow \mathrm{AB}=\frac{12}{\sin 122^{\circ}} \times \sin 13^{\circ} \\\\ & \mathrm{AB}=\frac{12}{0.8480} \times 0.2249=3.1833 \mathrm{~cm} \end{aligned} $$
In $\triangle \mathrm{BAE}$,
$$ \begin{aligned} & \frac{\mathrm{BE}}{\mathrm{AB}}=\sin 45^{\circ} \Rightarrow \mathrm{BE}=\mathrm{AB} \sin 45^{\circ} \\\\ & \mathrm{BE}=31833 \times \frac{1}{\sqrt{2}}=\frac{3.1833}{1.4142}=2.25 \mathrm{~cm} \end{aligned} $$
Since, $\mathrm{BE}=\mathrm{FG}$ is position of image $I_1$ at upper half of square, $h=2.25 \mathrm{~cm}$.
Similarly, in lower half of square image $I_2$ will be at a distance of $2.25 \mathrm{~cm}$.
$$ I_1 I_2=2 \times 2.25=4.50 \mathrm{~cm}=4.00 \mathrm{~cm} $$
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