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.
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:

$f_{e}$ = 1.7 cm

Explanation:

The magnification of the compound microscope is given by the product of the magnification of each lens

        M = M₀ $m_{e}$

        M = - L/f₀  25/$f_{e}$

Where f₀ and $f_{e}$ are the focal lengths of the lens and eyepiece, respectively, all values ​​in centimeters

In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece ($f_{e}$)

         $f_{e}$ = - L / f₀ 25 / M

Let's calculate

        $f_{e}$ = - 16 / 0.6 25 / (-400)

        $f_{e}$ = 1.67 cm

The minus sign in the magnification is because the image is inverted.

          $f_{e}$ = 1.7 cm