Chapter 25, Problem 30P

A microscope has an objective lens with a focal length of 16.22 mm and an eyepiece with a focal length of 9.50 mm. With the length of the barrel set at 29.0 cm, the diameter of a red blood cell’s image subtends an angle of 1.43 mrad with the eye. It the final image distance is 29.0 cm from the eyepiece, what is the actual diameter of the red blood cell? Hint: To solve this question, go back to basics and use the thin-lens equation.

To determine

The actual diameter of the red blood cell.

Answer to Problem 30P

The actual diameter of the red blood cell is $0.809\mu \text{m}$.

Explanation of Solution

The diameter of the final image is,

$h_e=r\theta$

• • $r$ is the distance between the image and the eye
  • $\theta$ is the angle subtended by the diameter of the image

Substitute $29.0\text{cm}$ for $r$ and $1.43\text{mrad}$ for $\theta$.

$\begin{array}{c}{h}_e=\left(29.0\text{cm}\right)\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\left(1.43\text{mrad}\right)\left(\frac{{10}^{-3}\text{rad}}{1\text{mrad}}\right)\\ =4.15\times {10}^{-4}\text{m}\end{array}$

The required object distance for the eyepiece is,

$p_e=\frac{q_e{f}_e}{q_e-f_e}$

• • $q_e$ is the image distance for the eyepiece
  • $f_e$ is the focal length of the eyepiece

Substitute $-29.0\text{cm}$ for $q_e$ and $0.950\text{cm}$ for $f_e$.

$\begin{array}{c}{p}_e=\frac{\left(-29.0\text{cm}\right)\left(0.950\text{cm}\right)}{-29.0\text{cm}-0.950\text{cm}}\\ =0.920\text{cm}\end{array}$

The magnification produced by the eyepiece is,

$\begin{array}{c}{M}_e=-\frac{q_e}{p_e}\\ =\frac{-\left(29.0\text{cm}\right)}{0.920\text{cm}}\\ =31.5\end{array}$

The image distance for the objective lens is,

$\begin{array}{c}{q}_1=L-p_e\\ =29.0\text{cm}-0.920\text{cm}\\ \text{=28}\text{.1}\text{cm}\end{array}$

The object distance for this lens is,

$p_1=\frac{q_1{f}_o}{q_1-f_o}$

Substitute $28.1\text{cm}$ for $q_1$ and $1.622\text{cm}$ for $f_o$.

$\begin{array}{c}{p}_1=\frac{\left(28.1\text{cm}\right)\left(1.622\text{cm}\right)}{28.1\text{cm}-1.622\text{cm}}\\ =1.72\text{cm}\end{array}$

The magnification of the objective lens is,

$\begin{array}{c}{M}_1=-\frac{q_1}{p_1}\\ =-\frac{28.1\text{cm}}{1.72\text{cm}}\\ =-16.3\end{array}$

The overall lateral magnification is,

$\begin{array}{c}{M}_{total}=M_1{M}_e\\ =\left(-16.3\right)\left(31.5\right)\\ =-513\end{array}$

The actual diameter of the red blood cell serving as the original object is,

$\begin{array}{c}h=\frac{h_e}{\left|M_{total}\right|}\\ =\frac{4.15\times {10}^{-4}\text{m}}{513}\\ =8.09\times {10}^{-7}\\ =0.809\mu \text{m}\end{array}$

Conclusion:

Thus, the actual diameter of the red blood cell is $\underset{\_}{0.809\mu \text{m}}$