When the object is self-luminous,the resolving power of a microscope is given by the expression

  • A
    $\frac{2\mu \sin \theta}{1.22 \lambda}$
  • B
    $\frac{\mu \sin \theta}{\lambda}$
  • C
    $\frac{2\mu \cos \theta}{1.22 \lambda}$
  • D
    $\frac{2\mu}{\lambda}$

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Similar Questions

The limit of resolution of a telescope is $3.0 \times 10^{-7} \text{ rad}$. Assuming that it is used to see the light of wavelength $525 \text{ nm}$ from a star, what should be the diameter of the objective (in $\text{ m}$)?

$A$ telescope with objective diameter $R$ is used to observe a distant star emitting light of wavelength $500 \text{ nm}$,at a resolution of $5 \times 10^{-7} \text{ radian}$. The value of $R$ is . . . . . . $\text{cm}$.

Assuming the human pupil to have a radius of $0.25 \ cm$ and a comfortable viewing distance of $25 \ cm$,the minimum separation between two objects that the human eye can resolve at $500 \ nm$ wavelength is nearly: (in $\mu m$)

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