Chapter 32, Problem 80P

(a)

To determine

The length of tube.

Explanation of Solution

Given:

The power of both lenses is $20\text{D}$ .

Distance between lenses is $30\text{cm}$ .

Formula used:

Write expression for length of tube.

  $L=D-f_o-f_e$

Here, $D$ is distance between lenses, $f_o$ is focal length of objective lens and $f_e$ is focal length of eye piece.

Substitute $\frac{1}{P_o}$ for $f_o$ and $\frac{1}{P_e}$ for $f_e$ in above expression.

  $L=D-\frac{1}{P_o}-\frac{1}{P_e}$  ........ (1)

Calculation:

Substitute $30\text{cm}$ for $D$ , $20\text{D}$ for $P_o$ and $20\text{D}$ for $P_e$ in equation (1).

  $\begin{array}{l}L=\left(30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)-\frac{1}{20}-\frac{1}{20}\\ L=0.20\text{m}\end{array}$

Conclusion:

Thus, the length of tube is $0.20\text{m}$ .

(b)

To determine

The lateral magnification of the objective lens.

Explanation of Solution

Given:

The power of both lenses is $20\text{D}$ .

Distance between lenses is $30\text{cm}$ .

Formula used:

Write expression for length of tube.

  $L=D-f_o-f_e$

Here, $D$ is distance between lenses, $f_o$ is focal length of objective lens and $f_e$ is focal length of eye piece.

Substitute $\frac{1}{P_o}$ for $f_o$ and $\frac{1}{P_e}$ for $f_e$ in above expression.

  $L=D-\frac{1}{P_o}-\frac{1}{P_e}$  ........ (1)

Write expression for lateral magnification of microscope.

  $m_o=-\frac{L}{f_o}$

Substitute $\frac{1}{P_o}$ for $f_o$ in above expression.

  $m_o=-L{P}_o$  ........ (2)

Calculation:

Substitute $30\text{cm}$ for $D$ , $20\text{D}$ for $P_o$ and $20\text{D}$ for $P_e$ in equation (1).

  $\begin{array}{l}L=\left(30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)-\frac{1}{20}-\frac{1}{20}\\ L=0.20\text{m}\end{array}$

Substitute $0.20\text{m}$ for $L$ and $20\text{D}$ for $P_o$ in equation (2).

  $\begin{array}{l}m=-\left(0.20\text{m}\right)\left(20\text{D}\right)\\ m=-4.0\end{array}$

Conclusion:

Thus, lateral magnification produced by objective lens is $-4.0$ .

(c)

To determine

The magnifying power of microscope.

Explanation of Solution

Given:

The power of both lenses is $20\text{D}$ .

Distance between lenses is $30\text{cm}$ .

Formula used:

Write expression for length of tube.

  $L=D-f_o-f_e$

Here, $D$ is distance between lenses, $f_o$ is focal length of objective lens and $f_e$ is focal length of eye piece.

Substitute $\frac{1}{P_o}$ for $f_o$ and $\frac{1}{P_e}$ for $f_e$ in above expression.

  $L=D-\frac{1}{P_o}-\frac{1}{P_e}$  ........ (1)

Write expression for lateral magnification of microscope.

  $m_o=-\frac{L}{f_o}$

Substitute $\frac{1}{P_o}$ for $f_o$ in above expression.

  $m_o=-L{P}_o$  ........ (2)

Write expression for magnifying power of microscope.

  $M=m_o\frac{x}{f_e}$

Here, $x$ is near point and $f_e$ is focal length of eyepiece.

Substitute $\frac{1}{P_e}$ for $f_e$ in above expression.

  $M=m_{o}x{P}_e$  ........ (3)

Calculation:

Substitute $30\text{cm}$ for $D$ , $20\text{D}$ for $P_o$ and $20\text{D}$ for $P_e$ in equation (1).

  $\begin{array}{l}L=\left(30\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)-\frac{1}{20}-\frac{1}{20}\\ L=0.20\text{m}\end{array}$

Substitute $0.20\text{m}$ for $L$ and $20\text{D}$ for $P_o$ in equation (2).

  $\begin{array}{l}{m}_o=-\left(0.20\text{m}\right)\left(20\text{D}\right)\\ m_o=-4.0\end{array}$

Substitute $-4.0$ for $m_o$ , $25\text{cm}$ for $x$ and $20\text{D}$ for $P_e$ in equation (3).

  $\begin{array}{l}M=\left(-4.0\right)\left(25\text{cm}\right)\left(\frac{1\text{m}}{100\text{cm}}\right)\left(20\text{D}\right)\\ M=-20\end{array}$

Conclusion:

Thus, magnifying power of microscope is $-20$ .

(d)

To determine

The distance of object from objective lens.

Explanation of Solution

Given:

The power of both lenses is $20\text{D}$ .

Distance between lenses is $30\text{cm}$ .

Formula used:

Write expression for lens equation for objective lens.

  $\frac{1}{f_o}=\frac{1}{v}+\frac{1}{u}$

Substitute $f_o+L$ for $v$ in above expression.

  $\frac{1}{f_o}=\frac{1}{u}+\frac{1}{f_o+L}$

Solve above expression for $u$ .

  $u=\frac{f_o\left(f_o+L\right)}{L}$  ........ (1)

Calculation:

Substitute $\frac{1}{20}\text{m}$ for $f_o$ and $20\text{cm}$ for $L$ in equation (1).

  $\begin{array}{l}u=\frac{\left(\frac{1}{20}\text{m}\right)\left(\frac{100\text{cm}}{1\text{m}}\right)\left(\left(\frac{1}{20}\text{m}\right)\left(\frac{100\text{cm}}{1\text{m}}\right)+20\text{cm}\right)}{20\text{cm}}\\ u=6.3\text{cm}\end{array}$

Conclusion:

Thus, the object is placed $6.3\text{cm}$ from the objective lens.