Why will a blue ray bend lesser than a red ray through a slit of the size a little bigger than the wavelength of the blue ray?

Don't think of bending. Think of diffraction like this: if you have a plane wave incident on a slit, then you can think about the space in the slit as being a line of infinitely many point sources that radiate in phase.

If you are looking straight down the slit, then all those point sources are in phase. There's not much unusual going on here.

However, if you move a bit to the side, then all those point sources aren't in phase. They are, really, but since they are not at equal distances to you, the radiation from each is delayed by a different amount. Depending on your position, the point sources interfere constructively or destructively, and this is what yields the diffraction pattern.

*If you look closely at this image, it appears it was generated by an approximation of four point sources in the slit.*

Now, the number of these point sources there are, and the maximum difference in phase between them, is a function of the size of the slit, obviously. If the slit is wider, then when viewed from some direction slightly off center, the phase difference from the left-most source and the right-most source will be greater, because the difference in distance between them is greater.

Compare a small slit:

To a bigger slit:

The significance of the size of the slit is apparent, right?

Well, changing the wavelength is equivalent to changing the size of the slit. If we make the slit bigger, and make the wavelength bigger by the same amount, then the difference in distance between the sources is greater, but the rate of change in the wave function is slower, so the phase difference between the two extremes of the slit is the same.

But, if we just make the wavelength smaller, and leave the slit the same, the rate of change in the wave function is faster, which is equivalent to making the slit bigger without changing the wavelength.

*Images from Wikipedia*

Unfortunately, I think you are speaking about what people commonly say is "Huygen's Principle", "In order to explain waves diffraction, it says that every point in a wave front behaves as a source, so the next wave front is the sum of all secondary waves produced by these points.", but this is not actually what Huygen's principle says.

Huygen's principle has to do with the propagation of light, which is electromagnetic waves, governed by Maxwell's equations. It can be shown that upon decoupling Maxwell's equations, one obtains spacetime wave equations of the form:

$u_{,t,t} = c^2 \left(u_{x,x} + u_{y,y} + u_{z,z}\right)$, (commas indicate partial derivatives) subject to the boundary conditions:
$u(\mathbf{x},0) = u(\mathbf{x}), \quad u_{,t}(\mathbf{x},0) = \psi(\mathbf{x})$.

The solution is given by D'Alembert's formula, but in the context of space-time wave equations, is known as Kirchhoff's formula or the Poisson formula, but it is the generalization of the Huygen-Fresnel equation, and is given by:

$$u(\mathbf{x},t_{0}) = \frac{1}{4\pi c^2 t_{0}} \iint_{S} \psi(\mathbf{x})dS + \left[\frac{1}{4 \pi c^2 t_{0}} \iint_{S} \phi(\mathbf{x}) dS\right]_{,t_{0}}.$$

You see from the solution that the point of Huygen's principle is to ensure causality of wave propagation. That is, as can be seen from the solution that $u(\mathbf{x}_{0},t_{0})$ depends on the boundary conditions on the spherical surface $S = \{ |\mathbf{x}-\mathbf{x}_{0}| = c t_{0} \}$, but not on the values **inside** the sphere! That is, the boundary conditions influence the solution only on the spherical surface $S$ of the light cone that is produced from this point.

This is precisely **Huygen's principle: Any solution of the spacetime wave equation travels at exactly the speed of light $c$**. So, as you can see Huygen's principle is independent of any specific slit/aperture configuration, it will apply in **any** situation where you can set up such boundary conditions for the spacetime wave equation!

## Best Answer

For larger objects the radio wave gets reflected. Compare this to a water wave hitting a wall. For smaller objects the radio gets diffracted. Compare this to a stick placed in the path of water wave. This stick bends the water wave which is similar to diffraction

A light wave consists of larger number of smaller waves. A mountain reflects most amount of these smaller waves but the tip is small compared to the wavelength. It diffracts the smaller waves which are incident on it

edit 1: http://www.acoustics.salford.ac.uk/feschools/waves/diffract.php