Question

A geneticist looks through a microscope to determine the phenotype of a fruit fly. The microscope is set to an overall magnification of 400x with an objective lens that has a focal length of 0.60 cm. The distance between the eyepiece and objective lenses is 16 cm. Find the focal length of the eyepiece lens assuming a near point of 25 cm (the closest an object can be and still be seen in focus).

Answer 1

Answer:

$f_e = 1.51 cm$

Explanation:

given.

magnification(m) = 400 x

focal length (f_0)= 0.6 cm

distance between eyepiece and lens (L)= 16 cm

Near point (N) = 25 cm

focal length of the eyepiece (f_e)= ?

using equation

$m = -\dfrac{L-f_e}{f_o}.\dfrac{N}{f_e}$

$400 = \dfrac{16-f_e}{0.6}.\dfrac{25}{f_e}$

$9.6 = \dfrac{16-f_e}{f_e}$

$9.6f_e = 16-f_e$

$10.6 f_e = 16$

$f_e = 1.51 cm$