Simple Microscope

The image formed by magnifying glass or simple microscope is virtual and erect object place between principal focus (f) and convex lens.

1. The normal district vision
2. The position of the lens is usually adjusted so that V is about 25cm, which is the shortest distance of distinct vision.

The lens formula relates the object distance (u), image distance (v), and focal length (f) of a lens:

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

Or rearranged:

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

• Object distance u is always taken as positive.
• If the image is virtual, then image distance v is negative.
• For a simple microscope, the image is virtual and located at the near point (typically v = -25 cm).

Given:

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

Substitute $v = -25 \, \text{cm}$:

$\frac{1}{u} = \frac{1}{f} - \frac{1}{25}$

Find common denominator:

$\frac{1}{u} = \frac{25 - f}{25f}$

Take the reciprocal:

$u = \frac{25f}{25 - f}$

Magnification is the ratio of the image size to the object size or the ratio of the image distance to the object distance:

$M = \frac{v}{u}$

From the formula of object distance:

$u = \frac{25f}{25 - f}$

Substitute into magnification formula:

$M = \frac{v}{\frac{25f}{25 - f}} = \frac{-25(25 - f)}{25f}$

$M = \frac{-25(25 - f)}{25f} = \frac{-(25 - f)}{f}$

Take magnitude (positive, since we are interested in size):

$M = \frac{25}{f} + 1$

Problem: A simple microscope has a lens of focal length f = 5 cm. It is used to observe divisions on a scale that are 0.5 mm in size. What is the magnified size of each division?

Solution:

Given:

• $f = 5 \, \text{cm}$
• Object size $h_o = 0.5 \, \text{mm}$

Step 1: Calculate magnification

$M = \frac{25}{f} + 1 = \frac{25}{5} + 1 = 5 + 1 = 6$

Step 2: Calculate image size

$h_i = M \times h_o = 6 \times 0.5 \, \text{mm} = 3.0 \, \text{mm}$

Answer: Each division will appear to be 3.0 mm in size when viewed through the simple microscope.

Parts of simple microscope