In a microscope of tube length 10 cm two convex lenses are arranged with focal length of 2 cm and 5

Question

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

Professor PC Thomas Classes · Answer

For a compound microscope in normal adjustment (with the final image formed at infinity), the total magnifying power is given by the product of the objective and eyepiece magnifications: $m = \left(\frac{L}{f_o}ight) \left(\frac{D}{f_e}ight)$. Given the tube length $L = 10 	ext{ cm}$, least distance of distinct vision $D = 25 	ext{ cm}$, $f_o = 2 	ext{ cm}$, and $f_e = 5 	ext{ cm}$. Substituting these values, $m = \left(\frac{10}{2}ight) 	imes \left(\frac{25}{5}ight) = 5 	imes 5 = 25$. We are given $m = (5)^k$, so $25 = 5^2 \implies k = 2$.