Chapter 25, Problem 45P

To determine

The largest and the smallest magnifications that can be obtained using the given lenses as objective and eyepiece.

Answer to Problem 45P

Solution:

The largest magnification that can be obtained is 620 x and the smallest magnification obtained is 25 x.

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.

The magnification of the microscope is the product of the magnifications of the objective and the eyepiece.

Using the thin lens formula, a relation between the object distance $d_o$ and the focal length of the objective $f_o$ is determined.

$\frac{1}{f_o}=\frac{1}{d_o}+\frac{1}{d_i}$

The magnification $m_o$ of the objective is given by,

$m_o=\frac{d_i}{d_o}$

Using the given values of the focal length, the maximum and the minimum values of $m_o$ are determined and these values when multiplied by the maximum and the minimum values of the magnification of the eye piece give the values of the maximum and the minimum possible values of magnification of the microscope.

$M=m_o\times m_e$

Given:

The values of the objective focal lengths $f_o=32\text{ mm, 15 mm and 3}\text{.9 mm}$

The values of the magnifications of the eyepiece $m_e=5\times ,15\times$

The distance from the objective at which the image is formed by it $d_i=f_o+\left(160\text{ mm}\right)$

Formula:

$\frac{1}{f_o}=\frac{1}{d_o}+\frac{1}{d_i}$

$m_o=\frac{d_i}{d_o}$

$M=m_o\times m_e$

Calculation:

Derive a relationship between the object distance and the focal length of the objective using the thin lens equation, by substituting $d_i=f_o+\left(160\text{ mm}\right)$ in the thin lens equation.

$\begin{array}{c}\frac{1}{d_o}=\frac{1}{f_o}-\frac{1}{d_i}\\ =\frac{1}{f_o}-\frac{1}{f_o+\left(160\text{ mm}\right)}\\ d_o=\frac{f\left[f_o+\left(160\text{ mm}\right)\right]}{160\text{ mm}}\end{array}$

Derive an expression for the magnification of the objective, using the expression,

$\begin{array}{c}{m}_o=\frac{d_i}{d_o}\\ =\frac{\left(f_o+\left(160\text{ mm}\right)\right)}{f_o\left[f_o+\left(160\text{ mm}\right)\right]/\left(160\text{ mm}\right)}\\ =\frac{\left(160\text{ mm}\right)}{f_o}\end{array}$

The objective has maximum magnification if the chosen focal length is the least. Of the given values, the smallest focal length is 3.9 mm.

$\begin{array}{c}{\left(m_o\right)}_{\max }=\frac{\left(160\text{ mm}\right)}{f_o}\\ =\frac{\left(160\text{ mm}\right)}{3.9\text{ mm}}\\ =41.03\end{array}$

The maximum magnification of the eyepiece is obtained using the lens with magnification $15\times$

The maximum total magnification that can be obtained is given by,

$\begin{array}{c}{M}_{\max }={\left(m_o\right)}_{\max }\times {\left(m_e\right)}_{\max }\\ =\left(41.03\right)\left(15\right)\\ =615\\ =620\times \left(2\text{ sf}\right)\end{array}$

The minimum magnification of the objective is obtained by using the value of maximum focal length.

$\begin{array}{c}{\left(m_o\right)}_{\min }=\frac{\left(160\text{ mm}\right)}{f_o}\\ =\frac{\left(160\text{ mm}\right)}{32\text{ mm}}\\ =5\end{array}$

The minimum magnification of the eyepiece is obtained using the lens with magnification $5\times$

The minimum total magnification that can be obtained is given by,

$\begin{array}{c}{M}_{\min }={\left(m_o\right)}_{\min }\times {\left(m_e\right)}_{\min }\\ =\left(5\right)\left(5\right)\\ =25\times \end{array}$