Question

A microscope consists of an objective with a focal length 2 mm and an eye piece with a focal length 40mm. The distance between the foci (which are between the lenses) of objective and eyepiece is 18 cm. The total magnification of the microscope is (Consider normal adjustment and take D $=$ 25 cm)

• A. 562.5
• B. 625
• C. 265
• D. 62.5

Answer 1

The correct option is A 562.5

The magnification of a microscope is given by,

$m=\frac{LD}{f_o{f}_e}$

where L is the distance between the foci, D is the least distance of distinct vision the other two terms are the focal length of the objective and the eye piece. Substituting the values given,

$m=\frac{18\times 25}{0.2\times 4}$

Thus, $m=562.5$