# Why is the shorter the wavelength, the smaller the object you can image? and vice-versa?

I know that the shorter the wavelength the smaller the object you can image clearly. Why wavelength matters in imaging something? How having big wavelength wont let u image smaller object, like if u have 400 nm wavelength and you are trying to image a 100 angstrom size of object, you wont be able to. That's what I know but I am not clear on the why behind it? Can anybody please explain the why

• The answer will depend on what you know of refraction and interference . Commented May 7 at 17:46
• @trula more so about diffraction Commented May 7 at 18:03

Here is a simple mental picture which may help.

Draw yourself a wave chain of sine waves going across the page. Now imagine inserting an object into the trough between two maxima in the curve. To get it to fit between the peaks requires it to be of a certain size. For sizes smaller than that it becomes possible for a sine wave to zing right by that object and not hit it at all because it "fits" easily between two maxima.

If that object is too big to fit between two maxima, then an incoming sine wave will "collide" with it and get scattered.

You can see it from the Rayleigh criterion or the resolution limit. For a classical optical microscope, the image of a point, $$\delta(x,y)$$, located in the object plane, isn't a point in the image plane. The image is the convolution of this point by a function called point spread function (PSF). This function is the impulse response of your microscope and is linked to the diffraction.

In microscopy, this PSF is an Airy function (or Airy disk in the 2D case) when using circular lenses. This function has a width (by width i mean the distance from the point of highest intensity to the first zero) directly related to the wavelength and the numerical aperture of your objective (and condenser to be precise).

Now, let's imagine two really close points in the object plane.The Rayleigh criterion is satisfied when the central maximum of the Airy Disk of one imaged point falls below the first minimum of the other.

This distance $$R$$ from the center of the Airy disk to the first minimum is given by $$$$R=\frac{0.61\lambda}{NA}$$$$

Where NA is the numerical aperture of your objective (a measure of the ray collection of your objective). From this formula, you see that if the wavelength increases, the width of your Airy disk increases. So as the wavelength gets larger, the image of a point gets larger too. So for two points really close in the object plane, you'll only see one bright spot in your image plane that is the superposition of your two Airy disks.

Here is a good tutorial to see the impact of $$\lambda$$ and NA on the image resolution: https://www.olympus-lifescience.com/en/microscope-resource/primer/java/imageformation/rayleighdisks/