Chapter 25, Problem 42P

To determine

The magnification of the given microscope

Answer to Problem 42P

Solution:

The magnification of the given microscope is found to be $520X$

Explanation of Solution

Microscopes are devices used to magnify tiny objects. The construction of the microscope is similar to that of the telescope. The objective produces a real and inverted image of the object, which falls between the focus and the optic centre of the eyepiece. The eyepiece produces an enlarged virtual image of the image formed by the objective. The final image formed is inverted and enlarged.

Given:

Focal length of the eyepiece $f_e=2.5\text{ cm}$

Barrel length $l=17\text{ cm}$

Focal length of the objective lens $f_o=0.33\text{ cm}$

Formula used:

For a relaxed eye, the image is formed at infinity. If the values of $f_e$ and $f_o$ are very less compared to the value of l, then, the magnification is given by,

$M=\frac{Nl}{f_e{f}_o}$

Here, N is the near point of the eye, for normal eye, N has a value of 25 cm.

Calculation:

Substitute the given values of N, l,$f_e$ and $f_o$ in the expression and simplify.

$\begin{array}{c}M=\frac{Nl}{f_e{f}_o}\\ =\frac{\left(25\text{ cm}\right)\left(17\text{ cm}\right)}{\left(2.5\text{ cm}\right)\left(0.33\text{ cm}\right)}\\ =515.1\\ =520X\end{array}$