0
$\begingroup$

enter image description here

I was not convinced by exactly how does this work,why do we need to take concentric spheres around point P and how do those spheres cast circular areas on the wavefront? Why is the radius OM1 and the first sphere originating from P not same?

$\endgroup$
1
  • $\begingroup$ Which book is this $\endgroup$
    – user167102
    Jan 8 '20 at 11:43
0
$\begingroup$

It's just commonly known diffraction in 2d. (https://en.wikipedia.org/wiki/Diffraction) The expression $d\sin(\theta_n)=n\lambda$ in polar symmetry.

$\endgroup$
0
$\begingroup$

Consider the wavefronts (orange) of a plane wave moving in from the left.

enter image description here

Each point on a wavefront acts as a secondary source - Huygen's construction.

To get to point $F$ the wavelets from point $X$ on the incident wavefront have to travel a distance $XF$.

The wavelets from a point $Y$ on the incident wavefront travel a distance $YF$ and so do not arrive in phase with the wavelets from point $X$.
At point $F$ there is a superposition of these two sets of wavelets.
As long as the path difference between the two sets of wavelets is less than $\frac \lambda 2$, where $\lambda$ is the wavelength of the incident waves the wavelets will superpose constructively ie combine together to give a greater amplitude of the wave at $F$.

If $AF$ and $A'F$ differ in distance by less than $\lambda 2$ then the wavelets which originate from the wavefront $AA'$ will arrive with a phase difference of less than $\frac \lambda 2$ and add together constructively.

Now consider the incident wavefront between $AB$ and consider a point $Z$ producing wavelets.
The distance $ZF$ is larger that $XF$ by at least $\frac \lambda 2$ and so the wavelet originate from $Z$ will be out of phase by greater than $\frac \lambda 2$ and so when they arrive at $F$ will superpose in a destructive manner reducing the amplitude of the resultant wave at $F$.
So that this not happen all waves which would produce a phase difference between $\frac \lambda 2\,(A)$ and $\lambda \, (B)$, and between $\frac \lambda 2\, (A')$ and $\lambda \,(B')$ are removed by making $AB$ and $A'B'$ opaque.

The next zone defined by $BC$ and $B'C'$ is made transparent because all the waves which arrive from that zone are between $\lambda$ and $\frac{3\lambda}{2}$ out of phase with the waves which came directly from $X$ and so will superpose constructively.

The end result is the constructive superposition of wave at $F$.

My diagram is just a section through the zone plate which has rotational symmetry about the line $XF$.

The radii of the circles which define the transparent and opaque zones do not increase in equal increments, just think about the geometry of the situation.

And finally a zone plate can be constructed with the transparent and opaque sections transposed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.