In a microscope of tube length 10 cm,two con | vedclass

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(A) For a compound microscope in normal adjustment,the objective lens forms the image at the focal point of the eyepiece. The tube length $L$ is the d

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$2$

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(A) For a compound microscope in normal adjustment,the objective lens forms the image at the focal point of the eyepiece. The tube length $L$ is the distance between the second focal point of the objective and the first focal point of the eyepiece. The magnification $M$ of a compound microscope in normal adjustment is given by the formula: $M = \left( \frac{L}{f_o} ight) 	imes \left( \frac{D}{f_e} ight)$. Given: $L = 10 \ cm$,$f_o = 2 \ cm$,$f_e = 5 \ cm$,and the least distance of distinct vision $D = 25 \ cm$. Substituting the values: $M = \left( \frac{10}{2} ight) 	imes \left( \frac{25}{5} ight) = 5 	imes 5 = 25$. We are given $M = (5)^k$. Therefore,$25 = (5)^k \implies 5^2 = 5^k$. Comparing the exponents,we get $k = 2$.

In a microscope of tube length $10 \ cm$,two convex lenses are arranged with focal lengths of $2 \ cm$ and $5 \ cm$. The total magnification obtained with this system for normal adjustment is $(5)^{k}$. The value of $k$ is . . . . . . .

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