Question

The focal lengths of the objective lens and the eyepiece of a microscope are 0.50 cm and 2.0 cm, respectively, and their separation is 6.0 cm when adjusted for minimum eyestrain for a person with a near point of 25.0 cm. If the microscope is focused on a small object, the magnitude of its final overall magnification is closest to

Answer 1

Answer:

- 103.7

Step-by-step explanation:

Given:

Focal length of the eyepiece, f = 2.0 cm

Focal length of the objective lens, f' = 0.50 cm

Separation for minimum eyestrain = 6,0 cm

Image distance, v = - 25 cm

Now, from the lens formula,

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$

here, u is the object distance

on substituting the respective values, we get

$\frac{1}{2}=\frac{1}{u}+\frac{1}{-25}$

or

u = 1.852 cm

also, the separation is adjusted for minimum eyestrain,

therefore, image distance for the objective lens, v' = 6 - 1.852 = 4.148 cm

Now, for the objective lens

using the lens formula, we get

$\frac{1}{0.5}=\frac{1}{u'}+\frac{1}{4.418}$

Here, u' is the distance between the physical object and objective lens

or

u' = 0.568 cm

Thus,

Magnification, m = $-\frac{vv'}{ff'}$

or

m =$-\frac{25\times4.418}{0.5\times2}$

or

m = - 103.7