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Optical Instruments — Physics STD 12 Science — Question

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

For the given compound microscope $f_0 = 1cm, f_e = 6cm, D = 24cm$
For the eye piece, $v_e= -24cm, f_e = 6cm$
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 9.8cm, the image distance for the objective lens must be $(9.8) - (4.8) = 5.0cm$
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 11.8cm,
$v_0 = 11.8 - 4.8 = 7.0cm, f_0 = 1cm$
$\Rightarrow\frac{1}{\text{u}_0}=\frac{1}{\text{v}_0}-\frac{1}{\text{f}_0}=\frac{1}{7}-\frac{1}{1}=-\frac{6}{7}$
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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