Verified Solution • 2025–26 Session moderate

QJEEAdv22Phc957: JEE Advanced 2022 Physics - Class 12 JEEAdv22Phc957 Solution: A rod of length makes an angle with the principal axis of a…

A rod of length 2 cm2 \mathrm{~cm} makes an angle 2π3rad\frac{2 \pi}{3} \mathrm{rad} with the principal axis of a thin convex lens. The lens has a focal length of 10 cm10 \mathrm{~cm} and is placed at a distance of 403 cm\frac{40}{3} \mathrm{~cm} from the object as shown in the figure. The height of the image is 30313 cm\frac{30 \sqrt{3}}{13} \mathrm{~cm} and the angle made by it with respect to the principal axis is α\alpha rad. The value of α\alpha is πnrad\frac{\pi}{n} r a d , where nn is __________ .

JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 19 English

Step-by-Step Verified Solution

JEE Advanced 2022 Paper 1 Online Physics - Geometrical Optics Question 19 English Explanation
hih0=vu⇒−303133=v−433⇒v1=43013 cm\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,

1v−1u=1f⇒1v=110−340⇒v=40 cm\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}

∴ x=40−43013=9013 cm\mathrm{x}=40-\frac{430}{13}=\frac{90}{13} \mathrm{~cm}

tan⁡α=3039013 N=13⇒α=30∘=π6\tan \alpha=\frac{\frac{30 \sqrt{3}}{\frac{90}{13}}}{\mathrm{~N}}=\frac{1}{\sqrt{3}} \Rightarrow \alpha=30^{\circ}=\frac{\pi}{6}

∴ N = 6

Related Learning Resources & Exam Preparation

JEE Advanced PYQ Question Bank
Browse year-wise authentic previous years' questions with solutions
Authentic CBT Mock Tests
Simulate official exam environment with live timer and percentile scoring