Chapter 39, Problem 27SP

In a compound microscope, the focal lengths of the objective and eyepiece are +0.50 cm and +2.0 cm, respectively. The instrument is focused on an object 0.52 cm from the objective lens. Compute the magnifying power of the microscope if the virtual image is viewed by the eye at a distance of 25 cm.

To determine

The magnifying power of the compound microscope, if the focal lengths of the objective and eye piece are $+0.50\text{ cm}$ and $+2.0\text{ cm}$, respectively.

Answer to Problem 27SP

Solution:

$3.4\times {10}^2$

Explanation of Solution

Given data:

The focal lengths of the objective and eye piece are $+0.50\text{ cm}$ and $+2.0\text{ cm}$, respectively.

The instrument is focused on an object $0.52\text{ cm}$ from the objective lens.

The virtual image is viewed by the eye at a distance of $25\text{cm}$.

Formula used:

Write the expression of focal length of objective.

$\frac{1}{f_o}=\frac{1}{s_o}+\frac{1}{s_i}$

Here, $f$ is the focal length of the correcting convex lens, $s_i$ is the image distance, and $s_o$ is the object distance.

Write the expression of magnification of compound microscope.

$M_A=\left(\frac{s_{io}}{s_{oo}}\right)\left(-\frac{s_{ie}}{s_{oe}}\right)$

Here, $M_A$ is the magnification of compound microscope, $s_{io}$ is the image distance from the objective lens to the intermediate image, $s_{oo}$ is the object distance from objective, $s_{ie}$ is the image distance from the eyepiece, and $s_{oe}$ is the object distance from the eyepiece.

Explanation:

Recall the expression of focal length of objective.

$\frac{1}{f_o}=\frac{1}{s_o}+\frac{1}{s_i}$

Substitute $+0.50\text{ cm}$ for $f_o$ and $0.52\text{ cm}$ for $s_o$

$\frac{1}{0.50\text{ cm}}=\frac{1}{0.52\text{ cm}}+\frac{1}{s_i}$

Solve for $s_i$.

$\begin{array}{c}\frac{1}{s_i}=\frac{1}{0.50\text{ cm}}-\frac{1}{0.52\text{ cm}}\\ =\frac{1}{0.50}-\frac{1}{0.52}{\text{ cm}}^{-1}\\ =\frac{0.02}{0.26}{\text{ cm}}^{-1}\\ s_i=13\text{ cm}\end{array}$

Recall the expression of focal length of eyepiece.

$\frac{1}{f_e}=\frac{1}{s_o}+\frac{1}{s_i}$

Substitute $+2.0\text{ cm}$ for $f_e$ and $-\text{25 cm}$ for $s_i$

$\frac{1}{2.0\text{ cm}}=\frac{1}{s_o}+\frac{1}{-25\text{ cm}}$

Solve for $s_i$.

$\begin{array}{c}\frac{1}{s_o}=\frac{1}{\text{2 cm}}+\frac{1}{\text{25 cm}}\\ =\frac{27}{50}{\text{ cm}}^{-1}\\ s_o=1.85\text{ cm}\end{array}$

The expression of magnification of compound microscope.

$\begin{array}{c}{M}_A=\left(\frac{s_{io}}{s_{oo}}\right)\left(-\frac{s_{ie}}{s_{oe}}\right)\\ =\left(\frac{13\text{ cm}}{0.52\text{ cm}}\right)\left(-\frac{-25\text{ cm}}{1.85\text{ cm}}\right)\\ =337.837\\ \simeq 3.378\times {10}^2\end{array}$

The magnification of compound microscope is,

$M_A=3.4\times {10}^2$

Conclusion:

Hence, the magnifying power of the microscope is $3.4\times {10}^2$.