The resolving power of an optical instrument | vedclass

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(A) The de-Broglie wavelength associated with an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.Thus,$\lambda \propto \frac{1}{\

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Increase the accelerating voltage.

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(A) The de-Broglie wavelength associated with an electron is given by $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2meV}}$.Thus,$\lambda \propto \frac{1}{\sqrt{V}}$.The resolving power of an optical instrument is inversely proportional to the wavelength,i.e.,$	ext{Resolving Power} \propto \frac{1}{\lambda}$.Substituting the relation for $\lambda$,we get $	ext{Resolving Power} \propto \sqrt{V}$.Therefore,if we increase the accelerating voltage $(V)$,the wavelength $(\lambda)$ decreases,which in turn increases the resolving power.

The resolving power of an optical instrument depends on the wavelength of the light used. An electron microscope uses a beam of electrons. Its resolving power can be increased if we:

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