Question

A compound microscope has an eyepiece of focal length 10 cm and an objective of focal length 5 cm. If an object is kept at a distance of 6 cm from the objective then magnification of the microscope for the final image at least distance of distinct vision of 25 cm is​

Answer 1

Answer:

Magnification produced is 17.5

Explanation:

fo = 5 cm

uo = 6 cm

1/vo -1/uo = 1/fo

1/vo = 1/5 - 1/6 = 1/30

vo = 30 cm

fe = 10 cm

D (Least distance)  = 25 cm

$m = \frac{Vo}{Uo} [ 1 + \frac{D}{fe} ]$

    = $\frac{30}{6} [ 1 + \frac{25}{10} ]$

    = 5 [ 1 + 2.5 ] = 5 x 3.5

                           = 17.5