Question
If $f _{0}=5\, cm , \lambda=6000\, \mathring A, a =1 \,cm$ for a microscope, then what will be its resolving power?
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Answer
The figure below represents the angle between the object and top of the lens of the microscope.
From the figure,
$\tan \theta \approx \sin \theta=\frac{a}{f}=\frac{1}{5}=0.2$
The resolving power of microscope is,
$R P=\frac{2 \mu \sin \theta}{1.22 \lambda}$
$=10.9 \times 10^{5} / m$
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