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Optical Instruments — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsOptical Instruments5 Marks
Question
The separation between the objective and the eyepiece of a compound microscope can be adjusted between $9.8\ cm$ to $11.8\ cm$. If the focal lengths of the objective and the eyepiece are $1.0\ cm$ and $6\ cm$ respectively, find the range of the magnifying power if the image is always needed at $24\ cm$ from the eye.
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Answer

For the given compound microscope $f_0 = 1\ cm, f_e = 6\ cm, D = 24\ cm$
For the eye piece$, v_e= -24\ cm, f_e = 6\ cm$
Now, $\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{f}_\text{e}}$
$\Rightarrow\frac{1}{\text{u}_\text{e}}=\frac{1}{\text{v}_\text{e}}-\frac{1}{\text{f}_\text{e}}$
$\Rightarrow-\Big[\frac{1}{24}+\frac{1}{6}\Big]=-\frac{5}{24}$
$\Rightarrow \text{u}_\text{e}=-4.8\ \text{cm}$
  1. When the separation between objective and eye piece is $9.8\ cm,$ the image distance for the objective lens must be $(9.8) - (4.8) = 5.0\ cm$
Now, $\frac{1}{\text{v}_0}-\frac{1}{\text{u}_0}=\frac{1}{\text{f}_0}$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{5}-\frac{1}{1}=-\frac{4}{5}$
$\Rightarrow\text{u}_0=-\frac{5}{4}=-1.25\text{cm}$
So, the magnifying power is given by,
$\text{m}=\frac{\text{v}_0}{\text{u}_\text{0}}\Big[1+\frac{\text{D}}{\text{f}}\Big]=\frac{-5}{-1.25}\Big[1+\frac{24}{6}\Big]=4\times5=20$
  1. When the separation is $11.8\ cm,$
$v_0 = 11.8 - 4.8 = 7.0\ cm, f_0 = 1\ cm$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{7}-\frac{1}{1}=-\frac{6}{7}$
So, $\text{m}=-\frac{\text{v}_0}{\text{u}_\text{o}}\Big[1+\frac{\text{D}}{\text{f}}\Big]$
$=\frac{-7}{-\Big(\frac{7}{6}\Big)}\Big[1+\frac{24}{6}\Big]=6\times5=30$
So, the range of magnifying power will be $20$ to $30.​​​​​​​$​​​​​​​

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