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 0.40 cm. The distance between the eyepiece and objective lenses is 12 cm. 1) 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). Do not neglect any values in your calculation. (Express your answer to two significant figures.)

Answer 1

Answer:

$f_{e}$  = 1.9 cm

Explanation:

The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective

    M = M₀ $m_{e}$

Where M₀ is the magnification of the objective and  $m_{e}$ is the magnification of the eyepiece.

The eyepiece is focused to the near vision point (d = 25 cm)

       $m_{e}$ = 25 /  $f_{e}$

The objective is focused on the distances of the tube (L)

     M₀ = -L / f₀

Substituting

     M = - L/f₀    25/$f_{e}$  

1) Let's look for the focal length of the eyepiece (faith)

     $f_{e}$  = - L 25 / f₀ M

     M = 400X = -400

     $f_{e}$  = - 12 25 /0.40 (-400)

     $f_{e}$  = 1.875 cm

Let's approximate two significant figures

    $f_{e}$  = 1.9 cm