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QJEEM23Ph227e: JEE Main 2023 Physics - Class 12 JEEM23Ph227e Solution: A thin cylindrical rod of length is placed horizontally on the…

A thin cylindrical rod of length 10 cm10 \mathrm{~cm} is placed horizontally on the principle axis of a concave mirror of focal length 20 cm20 \mathrm{~cm} . The rod is placed in a such a way that mid point of the rod is at 40 cm40 \mathrm{~cm} from the pole of mirror. The length of the image formed by the mirror will be x3 cm\frac{x}{3} \mathrm{~cm} . The value of xx is _____________.

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JEE Main 2023 (Online) 1st February Morning Shift Physics - Geometrical Optics Question 74 English Explanation

 A: \text { A: }

amp;1v+1u=1famp;⇒1v+1−45=1−20amp;⇒1v=145−120=4−9180=−136amp;⇒v=−36 cm\begin{aligned}& \frac{1}{v}+\frac{1}{u}=\frac{1}{f} \\\\& \Rightarrow \frac{1}{v}+\frac{1}{-45}=\frac{1}{-20} \\\\& \Rightarrow \frac{1}{v}=\frac{1}{45}-\frac{1}{20}=\frac{4-9}{180}=-\frac{1}{36} \\\\& \Rightarrow v=-36 \mathrm{~cm}\end{aligned}

B: 1v+1−35=1−20\frac{1}{v}+\frac{1}{-35}=\frac{1}{-20}

⇒1v=135−120=4−7140\Rightarrow \frac{1}{v}=\frac{1}{35}-\frac{1}{20}=\frac{4-7}{140}

⇒v=−1403\Rightarrow \quad v=-\frac{140}{3}

⇒\Rightarrow length of image =1403−36=323 cm=\frac{140}{3}-36=\frac{32}{3} \mathrm{~cm}

⇒x=32\Rightarrow x=32

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