We use a simple microscope to magnify an object. The microscope h… - Vidyadip

MCQ

We use a simple microscope to magnify an object. The microscope has a numerical aperture of $sin\ \alpha $ = $0.24$ . The object is so small that the resolving power of the microscope is totally employed. If the diameter of the eye's pupil is $d$ = $4.0\ mm$ , what is the minimum magnifying power of the microscope?

✓
$30$
B
$20$
C
$15$
D
$10$

✓

Answer

Correct option: A.

$30$

a
Minimum magnifying power $\Rightarrow$ Image is at $\infty$

$\mathrm{m}=\frac{\mathrm{D}}{\mathrm{f}}=\frac{\theta}{\theta_{0}}$

for microscope $\mathrm{d}_{\min }=\frac{0.61 \lambda}{\sin \alpha}$

$\frac{\mathrm{d}_{\min }}{\mathrm{D}}=\theta_{0}=\frac{0.61 \lambda}{\mathrm{D} \sin \alpha}$

for eye, $\theta_{\min }=\frac{1.22 \lambda}{\mathrm{d}}=\theta$

$\mathrm{m}_{\min }=\frac{\frac{1.22 \lambda}{\mathrm{d}}}{\frac{0.61 \lambda}{\mathrm{D} \sin \alpha}}=\frac{2 \mathrm{D} \sin \alpha}{\mathrm{d}}=30$

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