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Wave Optics — Physics STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 SciencePhysicsWave Optics5 Marks
Question
Explain the resolving power of an optical instrument. Write the formula for resolving power of telescope and microscope. Why is the resolving power of electron microscope more than that of optical microscope ?
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Answer

Resolving Power of an optical instrument : If two objects are very close to each other, then their diffraction pattern will also be very close and will overlap each other. If the overlapping in the diffraction pattern is small than the objects are seen separate when seen by optical instrument. The optical instrument is able to resolve the objects. It however, the overlapping is large, then the objects will not be seen as separate; they will be seen as one. In other words, the optical instrument is not resolving them. The power of an optical instument to produce distinctly separate images of two close objects is called the 'resolving power' of that instrument.
But, an optical instrument has a limit to form separate images of two objects placed very close to each other. That minimum distance between two objects when they can be seen as separate by an optical instrument is called the 'limit of resolution' of that instrument. Smaller the limit of resolution of an optical instrument, greater is said to be its resolving power.
$\therefore \quad$ Resolving power $\propto \frac{1}{\text { Resolvinglimit }}$
Resolving power of telescope :
R.P. $=\frac{d}{1.22 \lambda}$
where d = aperture (diameter) of objective of telescope.
$\lambda=$ wavelength of light used.
Resolving power of Microscope : The resolving power of a microscope is its ability to show two nearby closely lying objects as just distinguished and it is equal to the reciprocal of its limit of resolution.
Image
Resolving power of Electron Microscope :
Resolving power of a microscope is given by :
R.P $=\frac{2 n \sin \alpha}{1.22 \lambda}$.
From this formula it is obvious that it depends upon two factors, wavelength of light used $(\lambda)$ and numerical aperture $(n \sin \theta)$, the smaller the wavelength of light used the greater will be the resolving power. Now, when electrons are scattered or diffracted, they behave according to de-Broglie as light of wavelength given by :
$\lambda=h / m v$
where $h$ is Plank's constant, $m$ and $v$ are mass and velocity of electrons. The wavelength $\lambda$ as obtained above comes out to be about 5 x 10-10 cm. This is $10^5$ times smaller than that of visible light. Hence resolving power of electron microscope should be 100000 times greater than that of light microscope. But numerical aperture in electron microscope is usually smaller than that of the optical microscope on account of limitation set by lens aberration. This factor will therefore, reduce the resolving power. But even so the resolving power of electron microscope is far greater than that of optical microscope.

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