Solved Problem on Optical Instruments

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In a compound microscope, the focal length of the objective is 2 mm, and that of the ocular is 10 mm. The length of the tube that supports the lenses is 18 cm. The final image of the system is formed at 50 cm from the eyepiece. Determine:

a) The magnification of the objective and eyepiece;

b) The magnification of the microscope;

Problem data:

- Objective focal length:
*f*_{1}= 2 mm; - Eyepiece focal length:
*f*_{2}= 10 mm; - Distance from the image to the eyepiece:
*p'*_{2}= −50 cm; - Distance between lenses:
*d*= 18 cm.

Using the rule,

Using the rule,

Image

Again, a parallel ray passes through the image focus, this time from the eyepiece

These two rays do not determine an image on the observer's side. To determine the image it is necessary to extend these rays to the side of the object

Problem diagram:

Using the sign convention, on the incident light side, we have the positive abscissa for the real object,

Image

Solution

a) First, let us convert all units to centimeters (cm)

\[
\begin{gather}
f_{1}=2\;\text{mm}=2\times 10^{-1}\;\text{cm}=0.2\;\text{cm}\\[10pt]
f_{2}=10\;\text{mm}=10\times 10^{-1}\;\text{cm}=1\;\text{cm}
\end{gather}
\]

To find
\[
\begin{gather}
\bbox[#99CCFF,10px]
{\frac{1}{f}=\frac{1}{p}+\frac{1}{p'}}
\end{gather}
\]

applying to the second lens (eyepiece)
\[
\begin{gather}
\frac{1}{f_{2}}=\frac{1}{p_{2}}+\frac{1}{p'_{2}}\\[5pt]
\frac{1}{p_{2}}=\frac{1}{f_{2}}-\frac{1}{p'_{2}}\\[5pt]
\frac{1}{p_{2}}=\frac{1}{1}-\frac{1}{(-50)}\\[5pt]
\frac{1}{p_{2}}=1+\frac{1}{50}
\end{gather}
\]

multiplying and dividing by 50, the first term on the right-hand side of the equation
\[
\begin{gather}
\frac{1}{p_{2}}=\frac{50}{50}\times 1+\frac{1}{50}\\[5pt]
\frac{1}{p_{2}}=\frac{50+1}{50}\\[5pt]
\frac{1}{p_{2}}=\frac{51}{50}\\[5pt]
p_{2}=\frac{50}{51} \tag{I}\\[5pt]
p_{2}=0.98\;\text{cm}
\end{gather}
\]

the eyepiece magnification
\[
\begin{gather}
\bbox[#99CCFF,10px]
{M_{e}=\frac{i_{2}}{o_{2}}=-{\frac{p'_{2}}{p_{2}}}}
\end{gather}
\]

using for
\[
\begin{gather}
M_{e}=-{\frac{(-50)}{\dfrac{50}{51}}}\\[5pt]
M_{e}=\cancel{50}.\frac{51}{\cancel{50}}
\end{gather}
\]

\[
\begin{gather}
\bbox[#FFCCCC,10px]
{M_{e}=51}
\end{gather}
\]

The length of the tube will be the sum of the distance from the objective to the image,
\( i_{1}\equiv o_{2} \),
\[
\begin{gather}
d=p'_{1}+p_{2}\\[5pt]
18=p'_{1}+0.98\\[5pt]
p'_{1}=18-0.98\\[5pt]
p'_{1}=17.02\;\text{cm}
\end{gather}
\]

Applying the
\[
\begin{gather}
\frac{1}{f_{1}}=\frac{1}{p_{1}}+\frac{1}{p'_{1}}\\[5pt]
\frac{1}{p_{1}}=\frac{1}{f_{1}}-\frac{1}{p'_{1}}\\[5pt]
\frac{1}{p_{1}}=\frac{1}{0.2}-\frac{1}{17.02}
\end{gather}
\]

writing
\( 0.2=\frac{2}{10} \)
and
\( 17.02=\frac{1702}{100} \)
\[
\begin{gather}
\frac{1}{p_{1}}=\frac{1}{\dfrac{2}{10}}+\frac{1}{\dfrac{1702}{100}}\\[5pt]
\frac{1}{p_{1}}=\frac{10}{2}+\frac{100}{1702}
\end{gather}
\]

multiplying and dividing by 851, the first term on the right-hand side of the equation
\[
\begin{gather}
\frac{1}{p_{1}}=\frac{8510-100}{1702}\\[5pt]
\frac{1}{p_{1}}=\frac{8410}{1702}\\[5pt]
p_{1}=\frac{1702}{8410} \tag{II}\\[5pt]
p_{1}\approx 0.2\;\text{cm}
\end{gather}
\]

the magnification of objective
\[
\begin{gather}
\bbox[#99CCFF,10px]
{M_{o}=\frac{i_{1}}{o_{1}}=-{\frac{p'_{1}}{p_{1}}}}
\end{gather}
\]

using for
\[
\begin{gather}
M_{o}=-{\frac{\dfrac{1702}{100}}{\dfrac{1702}{8410}}}\\[5pt]
M_{o}=-{\frac{\cancel{1702}}{100}\times\frac{8410}{\cancel{1702}}}\\[5pt]
M_{o}=-{\frac{8410}{100}}
\end{gather}
\]

\[
\begin{gather}
\bbox[#FFCCCC,10px]
{M_{o}=-84.1}
\end{gather}
\]

b) The magnification of the microscope will be

\[
\begin{gather}
M=M_{o}.M_{e}\\[5pt]
M=(-84.1)\times 51
\end{gather}
\]

\[
\begin{gather}
\bbox[#FFCCCC,10px]
{M=-4289.1}
\end{gather}
\]

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Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .