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JEE Main 2020 (4 Sep, Shift 1) · Optical Instruments & the Eye

Optical Instruments & the Eye question from JEE Main 2020 (4 Sep, Shift 1)

Q33JEE Main 2020 · 4 SEP · SHIFT 1Medium

Optical Instruments & the Eye

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

Numerical answer type. Work it out, then open the answer.

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Answer: 6.25

Image at 25 cm: $M = \dfrac{L}{f_o}\left(1 + \dfrac{D}{f_e}\right)$ .
$100 = \dfrac{20}{1}\left(1 + \dfrac{25}{f_e}\right) \Rightarrow 1 + \dfrac{25}{f_e} = 5$ .
$f_e = \dfrac{25}{4} = 6.25$ cm.

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