Chapter 24, Problem 49P

(a)

To determine

The distance between the lenses.

Answer to Problem 49P

The distance between the lenses is $19.0\text{cm}$.

Explanation of Solution

From problem $48\text{P}$, image distance for the eyepiece is $25.0\text{ cm}$ and the image distance for the objective $16.5\text{cm}$ and focal length of eyepiece is $2.80\text{cm}$.

Write the lens equation for the eyepiece.

  $\frac{1}{p_{\text{e}}}+\frac{1}{q_{\text{e}}}=\frac{1}{f_{\text{e}}}$

Here, $p_{\text{e}}$ is the object distance for the eyepiece, $q_{\text{e}}$ is the image distance for the eyepiece, and $f_{\text{e}}$ is the focal length of the eyepiece.

Rearrange above equation to get $p_{\text{e}}$.

  $p_{\text{e}}={\left(\frac{1}{f_{\text{e}}}-\frac{1}{q_{\text{e}}}\right)}^{-1}$                                                                                                      (I)

Write the expression for the distance between the lenses.

  $d=q_{\text{o}}+p_{\text{e}}$                                                                                                              (II)

Here, $d$ is the distance between lenses and $q_{\text{o}}$ is the image distance for the objective.

Conclusion:

Substitute $2.80\text{cm}$ for $f_{\text{e}}$ and $-25.0\text{cm}$ for $q_{\text{e}}$ in equation (I) to get $p_{\text{e}}$.

  $\begin{array}{c}{p}_{\text{e}}={\left(\frac{1}{2.80\text{cm}}-\frac{1}{-25.0\text{cm}}\right)}^{-1}\\ =2.518\text{cm}\\ \text{=}2.52\text{cm}\end{array}$

Substitute $2.52\text{cm}$ for $p_{\text{e}}$ and $16.5\text{cm}$ for $q_{\text{o}}$ in equation (II) to get $d$.

  $\begin{array}{c}d=16.5\text{cm}+2.52\text{cm}\\ \text{=19}\text{.0}\text{cm}\end{array}$

Therefore, the distance between the lenses is $19.0\text{cm}$.

(b)

To determine

The angular magnification.

Answer to Problem 49P

The angular magnification of the microscope is $-318$.

Explanation of Solution

Write the angular magnification for the eyepiece.

  $M_{\text{e}}=\frac{N}{p_{\text{e}}}$                                                                                                                 (III)

Here, $M_{\text{e}}$ is the angular magnification of eyepiece and $N$ is the near point.

Write the expression for the transverse magnification of the objective.

  $m_{\text{o}}=-\frac{q_{\text{o}}}{p_{\text{o}}}$                                                                                                             (IV)

Here, $m_{\text{o}}$ is the transverse magnification of the objective and $p_{\text{o}}$ is the object distance for the objective lens.

Write the lens equation for objective lens.

  $\frac{1}{p_{\text{o}}}+\frac{1}{q_{\text{o}}}=\frac{1}{f_{\text{o}}}$

Here, $f_{\text{o}}$ is the focal length objective lens.

Rearrange above equation to get $\frac{1}{p_{\text{o}}}$.

  $\frac{1}{p_{\text{o}}}=\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}$                                                                                                          (V)

Substitute $\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}$ for $\frac{1}{p_{\text{o}}}$ in equation (IV) to get $m_{\text{o}}$.

  $m_{\text{o}}=-q_{\text{o}}\left(\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}\right)$

Write the expression for total magnification.

  $M_{\text{total}}=m_{\text{o}}{M}_{\text{e}}$                                                                                                         (VI)

Substitute $-q_{\text{o}}\left(\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}\right)$ for $m_{\text{o}}$ and $\frac{N}{p_{\text{e}}}$ for $M_{\text{e}}$ in equation (VI) to get $M_{\text{total}}$.

  $M_{\text{total}}=-q_{\text{o}}\left(\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}\right)\times \frac{N}{p_{\text{e}}}$                                                                                  (VII)

Conclusion:

Substitute $16.5\text{cm}$ for $q_{\text{o}}$, $5.00\text{mm}$ for $f_{\text{o}}$, $25.0\text{cm}$ for $N$ and $2.518\text{cm}$ for $p_{\text{e}}$ in equation (VII) to get $M_{\text{total}}$.

  $\begin{array}{c}{M}_{\text{total}}=-\left(16.5\text{cm}\right)\left(\frac{1}{5.00\text{mm}\frac{1\text{ cm}}{10\text{ mm}}}-\frac{1}{16.5\text{cm}}\right)\times \frac{25.0\text{cm}}{2.518\text{cm}}\\ =-318\end{array}$

Therefore, the angular magnification of the microscope is $-318$.

(c)

To determine

The distance from the objective at which the object should be placed to get the final image at viewer’s near point.

Answer to Problem 49P

The distance from the objective at which the object should be placed to get the final image at viewer’s near point is $5.16\text{mm}$.

Explanation of Solution

Rearrange equation (V) to get $p_0$.

  $p_o={\left(\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}\right)}^{-1}$                                                                                                 (VIII)

Conclusion:

Substitute $16.5\text{cm}$ for $q_{\text{o}}$ and $5.00\text{mm}$ for $f_{\text{o}}$ in equation (VIII) to get $p_o$.

  $\begin{array}{c}{p}_o={\left(\frac{1}{5.00\text{mm}}-\frac{1}{16.5\text{cm}\times \frac{10\text{mm}}{1\text{cm}}}\right)}^{-1}\\ =5.16\text{mm}\end{array}$

Therefore, the distance from the objective at which the object should be placed to get the final image at viewer’s near point is $5.16\text{mm}$.