Chapter 24, Problem 48P

(a)

To determine

The distance between the lenses.

Answer to Problem 48P

The distance between the lenses is $19.3\text{cm}$.

Explanation of Solution

Assume that the final image is appeared at infinity. This happens when image of the objective is formed at focal length of the eyepiece. Therefore, distance between the lenses is the sum of focal length of eyepiece and image distance of the objective.

Write the expression for the distance between the lenses.

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

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

Conclusion:

In problem it is given that, focal length of eyepiece is $2.80\text{cm}$ and image distance of object is $16.5\text{cm}$.

Substitute $2.80\text{cm}$ for $p_{\text{e}}$ and $16.5\text{cm}$ for $q_{\text{o}}$ in equation (I) to get $d$.

  $\begin{array}{c}d=16.5\text{cm}+2.80\text{cm}\\ \text{=19}\text{.3}\text{cm}\end{array}$

Therefore, the distance between the lenses is $19.3\text{cm}$.

(b)

To determine

The angular magnification.

Answer to Problem 48P

The angular magnification of the microscope is $-286$.

Explanation of Solution

Write the angular magnification for the eyepiece.

  $M_{\text{e}}=\frac{N}{f_{\text{e}}}$                                                                                                           (II)

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

Write the expression for the transverse magnification.

  $m_{\text{o}}=-\frac{L}{f_{\text{o}}}$                                                                                                         (III)

Here, $L$ is the tube length of the microscope and $f_0$ is the focal length of objective.

Write the expression for the tube length of microscope.

  $L=q_0-f_0$

Substitute $q_{\text{o}}-f_{\text{o}}$ for $L$ in equation (III) to get $m_{\text{o}}$.

  $m_{\text{o}}=\frac{q_{\text{o}}-f_{\text{o}}}{f_{\text{o}}}$

Write the expression for total magnification.

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

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

  $M_{\text{total}}=-\frac{q_{\text{o}}-f_{\text{o}}}{f_{\text{o}}}\times \frac{N}{f_{\text{e}}}$                                                                                           (V)

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 $2.80\text{cm}$ for $f_{\text{e}}$ in equation (V) to get $M_{\text{total}}$.

  $\begin{array}{l}{M}_{\text{total}}=-\frac{16.5\text{cm}-\left(5.00\text{mm}\times \frac{1\text{cm}}{100\text{mm}}\right)}{5.00\text{mm}\times \frac{1\text{cm}}{100\text{mm}}}\times \frac{25.0\text{cm}}{2.80\text{cm}}\\ =-286\end{array}$

Therefore, the angular magnification of the microscope is $-286$.

(c)

To determine

The distance at which the object should be placed.

Answer to Problem 48P

The distance at which the object should be placed is $5.16\text{mm}$.

Explanation of Solution

Write mirror equation of objective.

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

Rearrange above equation to get $p_0$.

  $p_{\text{0}}={\left(\frac{1}{f_{\text{o}}}-\frac{1}{q_{\text{o}}}\right)}^{-1}$                                                                                            (VI)

Conclusion:

Substitute $16.5\text{cm}$ for $q_{\text{o}}$ and $5.00\text{mm}$ for $f_{\text{o}}$ in equation (VI) to get $p_{\text{0}}$.

  $\begin{array}{c}{p}_{\text{0}}={\left(\frac{1}{5.00\text{mm}}-\frac{1}{16.5\text{cm}\times \frac{10\text{mm}}{1\text{cm}}}\right)}^{-1}\\ =5.16\text{mm}\end{array}$

Therefore, the distance at which the object should be placed is $5.16\text{mm}$.