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Q.

The value of numerical aperture of the objective lens of a microscope is 1.25. If light of wavelength $5000 \overset{o}{A}$ is used then minimum separation between two points, to be seen as distinct is

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a

$0.24 μ m$

b

$0.48 μ m$

c

$0.12 μ m$

d

$0.38 μ m$

answer is A.

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Detailed Solution

Numerical aperture $= \frac{0.61 λ}{d}$
$\Rightarrow d = \frac{0.61 \times 0.5 μ m}{1.25} = 0.24 μ m$

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