JEE Advance - Physics (2022 - Paper 1 Online - No. 4)
A rod of length $2 \mathrm{~cm}$ makes an angle $\frac{2 \pi}{3} \mathrm{rad}$ with the principal axis of a thin convex lens. The lens has a focal length of $10 \mathrm{~cm}$ and is placed at a distance of $\frac{40}{3} \mathrm{~cm}$ from the object as shown in the figure. The height of the image is $\frac{30 \sqrt{3}}{13} \mathrm{~cm}$ and the angle made by it with respect to the principal axis is $\alpha$ rad. The value of $\alpha$ is $\frac{\pi}{n} r a d$, where $n$ is __________ .
Answer
6
Explanation
$\frac{\mathrm{h}_{\mathrm{i}}}{\mathrm{h}_0}=\frac{\mathrm{v}}{\mathrm{u}} \Rightarrow \frac{-\frac{30 \sqrt{3}}{13}}{\sqrt{3}}=\frac{\mathrm{v}}{-\frac{43}{3}} \Rightarrow \mathrm{v}_1=\frac{430}{13} \mathrm{~cm}$
Now using,
$ \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \Rightarrow \frac{1}{\mathrm{v}}=\frac{1}{10}-\frac{3}{40} \Rightarrow \mathrm{v}=40 \mathrm{~cm}$
$$ \therefore $$ $ \mathrm{x}=40-\frac{430}{13}=\frac{90}{13} \mathrm{~cm}$
$\tan \alpha=\frac{\frac{30 \sqrt{3}}{\frac{90}{13}}}{\mathrm{~N}}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=30^{\circ}=\frac{\pi}{6}$
$$ \therefore $$ N = 6
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