A compound microscope uses an objective lens of focal length 4 cm and eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also calculate the length of the microscope

Explanation

It is given that focal length of objective lens `f_(0)= + 4 cm` focal length of eyepiece lens `f_e = +10 cm` and `u_(0) = - 6 cm`
Let the final image is formed at the least distance of distinct vision (D = 25 cm).
For object lens `1/v_(0) -1/u_(0) =1/f_(0) rArr 1/v_(0) + 1/f_(0) = -1/6 + 1/4 =(-2+3)/12 = 1/12`
`therefore v_(0) = + 12 cm`
Again for eyepiece lens `v_e = - D = - 25 cm`, hence
`1/v_(e) -1/u_(e) =1/f_(e) rArr 1/u_(e) -1/_(f_(e)) =1/(-25) -1/(10) = (-2-5)/50 = -7/50`
or `u_(e) = -50/7 =-7.4 cm`
Length of the microscope tube `L = |v_(0)| + |u_(e)| = 12 + 7.4 = 19.4 cm`
and magnifying the power `m=m_(0) xx m_(e) =(v_(0)/u_(0)) xx(v_(e)/v_(f)) =12/(-6) xx (-25)/(-7.4) = -7`