# The value of numerical aperature of the objective lens of a microscope is 1.25. If light of wavelength 5000 A is used, the minimum seperation between two points, to be seen as distinct, will be : Option 1) $0.24 \mu m$ Option 2) $0.38\mu m$ Option 3) $0.12\mu m$ Option 4) $0.48 \mu m$

Resolving power of microscope -

$R= \frac{2\mu \sin \Theta }{\lambda }$

$\mu = Refractive\: index$

$\lambda = wavelength \: of \: light \: used$

-

Minimum distance between two lines$\varepsilon =0.61\times \frac{\lambda}{N.A.}$

where N.A is numerical opperature

$\Rightarrow \varepsilon =0.61\times \frac{5000\times10^{-10}}{1.25}$

$\Rightarrow \varepsilon =0.24\mu m$

Option 1)

$0.24 \mu m$

Option 2)

$0.38\mu m$

Option 3)

$0.12\mu m$

Option 4)

$0.48 \mu m$

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