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Ray Optics

JEE Main level, numerical answer, medium

281 of 295

Question 281 of 295

In a compound microscope, the magnified virtual image is formed at a distance of $25\text{cm}$ from the eye-piece. The focal length of its objective lens is $1\text{cm}$ . If the magnification is $100$ and the tube length of the microscope is $20\text{cm}$ , then the focal length of the eye-piece lens (in $\text{cm}$ ) is ______

Show answer and solution

Answer: 6.25

M=\dfrac{{v}_{0}}{{u}_{0}}\left(1+\dfrac{D}{{f}_{e}}\right)

M=\dfrac{L}{{f}_{0}}\left(1+\dfrac{D}{{f}_{e}}\right)

100=\dfrac{20}{\left(1\right)}\left(1+\dfrac{25}{{f}_{e}}\right)

;

5=1+\dfrac{25}{{f}_{e}}

4=\dfrac{25}{{f}_{e}}

;

{f}_{e}=\dfrac{25}{4}=6.25cm

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JEE Main 2025 Apr 2 Shift 1 question paper

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