MCQ
If $f _{0}=5\, cm , \lambda=6000\, \mathring A, a =1 \,cm$ for a microscope, then what will be its resolving power?
- A$11.9 \times 10^{5} / m$
- ✓$10.9 \times 10^{5} / m$
- C$10.9 \times 10^{5} / m$
- D$10.9 \times 10^{3} / m$
✓
Answer
Correct option: B.
$10.9 \times 10^{5} / m$
b
The figure below represents the angle between the object and top of the lens of the microscope.
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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