Use of shorter wavelengths in microscope?

Question

My textbook has a chapter named optical instruments; in it it says that to build more powerful microscopes we need more energy to reach shorter wavelengths. My question is why do we need shorter wavelengths to observe say a surface of a metal?

Answer 1

The gaps you want to see between need to be larger than the thing you're using to see them

Think of it a little like those pin art toys, the smaller the pin the more definition you will be able to see. It is a similar process with light, the smaller the wavelength the higher the definition.

Answer 2

The resolving power (ability to distinguish two objects close together) of a microscope is limited by diffraction of the waves used to illuminate the object.
The shorter the wavelength of the wave the better is the resolving power, so microscope which use ultra violet light have a greater resolving power then those that use visible light.
Electron microscopes use accelerated electron beams where the wavelength of the electrons can be 100,000 smaller than that of visible light thus increasing the resolving power of such a microscope by a great deal - enough to resolve molecules.

The reference to energy was because as the energy of photons or electrons increases the wavelength decreases.

Answer 3

To state Farcher's Answer in somewhat different language: when information is encoded on an imaging wavefront by a variation in scattering / transmission, if that variation is faster than a wavelength, then the imprint cannot propagate - it leads to evanescent waves.

Imagine such imprint on a wavefront. To understand how the optical field will propagate, we decompose it, through Fourier analysis, into a system of plane waves, propagate the plane waves to wherever our image forming device is and reassemble the field at the imaging device through the inverse Fourier transform to calculate how the field has propagated. I talk about this in more detail in this answer here.

So we do our Fourier analysis: if there is variation over distances smaller than a wavelength, this means that there will be many plane waves with transverse wavenumbers $k_\perp$ greater than the wave's wavenumber $k$. That means the component of the wavenumber in the direction of propagation is $\sqrt{k^2-k_\perp^2}$, which is imaginary. The intensity of such plane wave components thus decay exponentially with distance from where the imprint is made. Imprint information in Fourier components corresponding the faster than wavelength variation are therefore nonpropagating and cannot reach the imaging device.