Question
On what factors does the resolving power of a microscope depend? How can it be increased?
✓
Answer
Resolving power of a microscope
$
=\frac{2 n \sin a }{\lambda}=\frac{2 NA }{\lambda}
$
where, $\alpha \equiv$ the half angle of the angular separation between the objects at the objective lens. $n \equiv$ the refractive index of the medium between the object and the objective, $\lambda \equiv$ the wavelength of the light used to illuminate the object, $N A=n \sin \alpha=$ the numerical aperture of the objective.
Thus, the resolving power of a microscope depends directly on the NA and inversely on $\lambda$. The resolving power is increased by:
1. increasing the numerical aperture using oil-immersion objective.
2. illuminating the object with smaller wavelength radiation. But our eyes are not very sensitive to the shorter wavelength blue end of the visible spectrum. Hence, ultraviolet radiation is used for illumination with quartz lenses, but then photographs must be taken to examine the image.
$
=\frac{2 n \sin a }{\lambda}=\frac{2 NA }{\lambda}
$
where, $\alpha \equiv$ the half angle of the angular separation between the objects at the objective lens. $n \equiv$ the refractive index of the medium between the object and the objective, $\lambda \equiv$ the wavelength of the light used to illuminate the object, $N A=n \sin \alpha=$ the numerical aperture of the objective.
Thus, the resolving power of a microscope depends directly on the NA and inversely on $\lambda$. The resolving power is increased by:
1. increasing the numerical aperture using oil-immersion objective.
2. illuminating the object with smaller wavelength radiation. But our eyes are not very sensitive to the shorter wavelength blue end of the visible spectrum. Hence, ultraviolet radiation is used for illumination with quartz lenses, but then photographs must be taken to examine the image.
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