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

 

Answers (1)

 

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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