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(C) The magnifying power of a compound microscope when the final image is formed at infinity is given by the formula:$M = \frac{L \cdot D}{f_o \cdot f

Vedclass · Accepted Answer

$2.5$

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(C) The magnifying power of a compound microscope when the final image is formed at infinity is given by the formula:$M = \frac{L \cdot D}{f_o \cdot f_e}$Where:$M = 400$ (Magnifying power)$L = 20\, cm$ (Tube length)$D = 25\, cm$ (Least distance of distinct vision)$f_o = 5\, mm = 0.5\, cm$ (Focal length of objective)$f_e = ?$ (Focal length of eye-piece)Substituting the values into the formula:$400 = \frac{20 	imes 25}{0.5 	imes f_e}$$400 = \frac{500}{0.5 	imes f_e}$$f_e = \frac{500}{400 	imes 0.5} = \frac{500}{200} = 2.5\, cm$Thus,the focal length of the eye-piece is $2.5\, cm$.

The magnifying power of a microscope with an objective of $5\, mm$ focal length is $400$. The length of its tube is $20\, cm$. Then the focal length of the eye-piece is.....$cm$

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