Question

The distance between the object to be viewed and the eyepiece of a compound microscope is 25.0 cm. The focal length of its objective lens is 0.200 cm and the eyepiece has a focal length of 2.60 cm. What is the magnitude of the total magnification of the microscope when used by the person of normal eyesight if it is designed for minimum eyestrain?

Answer 1

Answer:

The magnitude of the total magnification of the microscope is 1201.9.

Explanation:

Given that,

Object distance = 25.0 cm

Focal length of its objective lens = 0.200 cm

Focal length of eyepiece = 2.60 cm

We need to calculate the total magnification of the microscope

Using formula for microscope

$M=-\dfrac{L}{f_{o}}(\dfrac{25}{f_{e}})$

Where, L = object distance

$f_{o}$=Focal length of its objective lens

$f_{e}$=Focal length of eyepiece

Put the value into the formula

$M=\dfrac{25.0}{0.200}(\dfrac{25}{2.60})$

$M=-1201.9$

The magnitude of the total magnification of the microscope

$|M|=1201.9$

Hence, The magnitude of the total magnification of the microscope is 1201.9.