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QJEEAdv20Ph7a65: JEE Advanced 2020 Physics - Class 12 JEEAdv20Ph7a65 Solution: A large square container with thin transparent vertical walls…

A large square container with thin transparent vertical walls and filled with water (refractive index 43{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 ;

JEE Advanced 2020 Paper 2 Offline Physics - Geometrical Optics Question 51 English

Step-by-Step Verified Solution

Consider ray of light from object O\mathrm{O} moves along OB\mathrm{OB} at an angle of incidence ii and refracted through air parallel to line of sight, that is, ∠r=45∘\angle r=45^{\circ} . Another ray slight above OA\mathrm{OA} is refracted along AC\mathrm{AC} and image appears to be formed at I1I_1 having height hh .

JEE Advanced 2020 Paper 2 Offline Physics - Geometrical Optics Question 51 English Explanation
At point B; from Snell's law

amp;sin⁡isin⁡45=14/3⇒sin⁡i=34sin⁡45∘amp;⇒sin⁡=34×12=34×1.4142=0.5303\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∘i=32.03^{\circ} .

In △OAB,∠B=(90∘+i)=(90∘+32.03∘)⇒122.03∘\triangle \mathrm{OAB}, \quad \angle B=\left(90^{\circ}+i\right)=\left(90^{\circ}+32.03^{\circ}\right) \Rightarrow 122.03^{\circ} .

amp;∠A=45∘amp;∠O=(180−45∘−122.03∘)≅13∘\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

amp;12sin⁡122∘=ABsin⁡13∘⇒AB=12sin⁡122∘×sin⁡13∘amp;AB=120.8480×0.2249=3.1833 cm\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 △BAE\triangle \mathrm{BAE} ,

amp;BEAB=sin⁡45∘⇒BE=ABsin⁡45∘amp;BE=31833×12=3.18331.4142=2.25 cm\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, BE=FG\mathrm{BE}=\mathrm{FG} is position of image I1I_1 at upper half of square, h=2.25 cmh=2.25 \mathrm{~cm} .

Similarly, in lower half of square image I2I_2 will be at a distance of 2.25 cm2.25 \mathrm{~cm} .

I1I2=2×2.25=4.50 cm=4.00 cmI_1 I_2=2 \times 2.25=4.50 \mathrm{~cm}=4.00 \mathrm{~cm}

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