Question

Diffraction if slit size is less than wavelength and why diffraction is most pronounced when slit length is comparable to wavelength?

From mathemetical treatment of diffraction we know that for first minima
`dsin(theta)=mlamda` So when slit size is less than the wavelength `sintheta=(mlamda)/d` which is (greater)>than 1 as m=1 and d less than lamda but sing greater than 1 is not possible

Answer 1

We don’t tend to use the equation to predict whether or not diffraction will occur, only to predict where the bright fringes will appear.

Explanation:

The explanation for the effect is that when the gap becomes less than a wavelength the diffraction disappears rapidly is simply that the wave cannot transmit energy through the gap. The wave responds as if there were no gap at all and it is predominantly reflected.

You also need to be clear the the ‘d’ in the equation you quote refers to the distance between the slits, not the aperture (width) of a slit itself. The is generally referred to as ‘a’ in textbooks in the UK and is used to calculate diffraction from a single slit (called Fraunhofer diffraction.) See here (http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/sinslit.html) for some further details.

Just noticed I hadn’t addressed the second part of your question very effectively, for this you need to understand the Huygens (wavelet) principle. Effectively each wave can be represented as a multitude of tiny wavelets each in phase. On passing through a gap these wavelets either reinforce (interfere constructively) or dissipate (interfere destructively) depending on the relative sizes of the wavelength and gap.

It is a long time since I studied this, so before I get bogged down, I’ll give you a reference to follow up: https://www.reddit.com/r/askscience/comments/3n9btn/why_does_the_wavelength_of_a_wave_affect/

Lucasvp describes it well in that post, but you may need to look at the underlying principle (Huygen’s) first which can be found here: https://en.m.wikipedia.org/wiki/Diffraction