If we need a magnification of 375 from a comp | vedclass

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(A) The magnification $M$ of a compound microscope is given by the formula $M = \frac{L}{f_0} \left(1 + \frac{d}{f_e}\right)$,where $L$ is the tube le

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$22$

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(A) The magnification $M$ of a compound microscope is given by the formula $M = \frac{L}{f_0} \left(1 + \frac{d}{f_e}\right)$,where $L$ is the tube length,$f_0$ is the focal length of the objective,$f_e$ is the focal length of the eyepiece,and $d$ is the least distance of distinct vision $(d = 250\; mm)$.Given: $M = 375$,$L = 150\; mm$,$f_0 = 5\; mm$,$d = 250\; mm$.Substituting the values into the formula:$375 = \frac{150}{5} \left(1 + \frac{250}{f_e}\right)$$375 = 30 \left(1 + \frac{250}{f_e}\right)$$12.5 = 1 + \frac{250}{f_e}$$11.5 = \frac{250}{f_e}$$f_e = \frac{250}{11.5} \approx 21.74\; mm$.Rounding to the nearest integer,$f_e \approx 22\; mm$.

If we need a magnification of $375$ from a compound microscope of tube length $150\; mm$ and an objective of focal length $5\; mm$,the focal length of the eye-piece should be close to.....$mm$.

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