91Ó°ÊÓ

The focal length of the objective,$f_{ob}=4\text{cm}=\text{0.04}\text{m}$ .

The focal length of the eyepiece,$f_{ey}=8\text{cm}=\text{0.08}\text{m}$.

Distance between the lenses,$L=25\text{cm}=\text{0.25}\text{m}$ .

From the fig.34.20, write an expression for the tube length. Inserting the given values will give its value. Using the lens law, find the distance between the objective and the object. Then using the corresponding formulae, find the magnification of the objective, angular magnification of the eyepiece, and overall magnification of the compound microscope.

Formula are as follows:

• Tube length,$\mathit{s}\mathbf{=}\mathit{L}\mathbf{-}{\mathit{f}}_{\mathbf{o}\mathbf{b}}\mathbf{-}{\mathit{f}}_{\mathbf{e}\mathbf{y}}$.
  
• Thin lens formula, $\frac{\mathbf{1}}{{\mathbf{f}}_{\mathbf{0}\mathbf{b}}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{p}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{i}}$.
  
• Magnification, $\mathit{m}\mathbf{=}\mathbf{-}\frac{\mathbf{i}}{\mathbf{p}}$.
  
• Overall magnification,$\mathit{M}\mathbf{=}\mathit{m}{\mathit{m}}_{\mathbf{θ}}$.
• Angular magnification, ${\mathit{m}}_{\mathbf{θ}}\mathbf{=}\frac{\mathbf{25}}{{\mathbf{f}}_{\mathbf{e}\mathbf{y}}}$.

Where m is the magnification, p is the pole, f is the focal length, and i is the image distance, m_θis theangular magnification,s is the tube length.

(a)

From figure 34.20,

$s=L-f_{ob}-f_{ey}$

Inserting the given values,

$s=0.25-0.04-0.08$

role="math" localid="1663019531134" $s=0.13\text{m}=\text{13}\text{cm}$

(b)

To calculate the distance between the objective and the object (p) needs the image distance I.

From the figure, write,

$$\begin{gathered} i=f_{ob}+s \\ i=0.04+0.13 \\ i=0.17\text{m} \end{gathered}$$

The lens formula is given by,

$\begin{array}{rcl}\frac{1}{f_{0b}}& =& \frac{1}{p}+\frac{1}{i}\\ \frac{1}{0.04}& =& \frac{1}{p}+\frac{1}{0.17}\\ \frac{1}{p}& =& \frac{1}{0.04}-\frac{1}{0.17}\\ p& =& 0.0523\text{m}=\text{5.23}\text{cm}\\ & & \end{array}$

Therefore, the distance between the objective and the object is 5.23 cm.

(c)

Find the magnification of the objective using the values of i and p as follows:

$$\begin{gathered} m=-\frac{i}{p} \\ m=-\frac{0.17}{0.0523} \\ m=-3.25 \end{gathered}$$

Therefore, the magnification of the objective is -3.25.

(d)

The angular magnification of the eyepiece can be calculated as,

$$\begin{gathered} m_{\mathrm{θ}}=\frac{\text{25}\text{cm}}{f_{ey}} \\ {m}_{\mathrm{θ}}=\frac{\text{25}\text{cm}}{\text{8}\text{cm}} \\ {m}_{\mathrm{θ}}=3.13 \end{gathered}$$

Therefore, the angular magnification of the eyepiece is 3.13.

(e)

Overall magnification of the microscope is,

$\begin{array}{rcl}M& =& m{m}_{\mathrm{θ}}\\ M& =& -3.25×3.13\\ M& =& -10.1725\\ & & ~-10.2\\ & & \end{array}$

The tube length, the distance between the objective and object, magnification of the objective, angular magnification of the eyepiece, and overall magnification of the compound microscope can be found using the corresponding formulae.