Physics - Fresnel’s class Diffraction - Prof. Bhandri, Study notes of Engineering Physics

Detail Summery about Fresnel’s class Diffraction, Fresnel’s half - period zones , Diffraction by an opaque Disc – The Poisson Spot, The zone –plate, mutually interfere, Huygens – Fresnels principle.

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2010/2011

Uploaded on 09/03/2011

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Fresnel’s class Diffraction
To explain Fresnel’s class difrection we use
Huygens – Fresnels principle according to which:
Each point on a wavefront is a source of
secondary disturbance and the secondary wavelets
emanating from different points mutually interfere.
We know that the incidence beam will undergo
diffraction divergence and the angular spreading will be
given by
Δθ ~ λ/2a
pf3
pf4
pf5
pf8
pf9
pfa
pfd

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Fresnel’s class Diffraction

To explain Fresnel’s class difrection we use Huygens – Fresnels principle according to which: Each point on a wavefront is a source of secondary disturbance and the secondary wavelets emanating from different points mutually interfere. We know that the incidence beam will undergo diffraction divergence and the angular spreading will be given by Δθ ~ λ/2a

If a >> λ, the intensity at a point R will be negligible; on the other hand , if a ~ λ there will be almost uniform spreading out of the beam resulting in an uniform illumination of the screen.

We make some construction as shown in the fig.

  1. Let PO = d and with point P we draw spheres of radii d+ λ/2, d+ 2λ/2, d+ 3λ/2, …….., these spheres will intersect WW’ in the circles. The radius of the nth circle will be rn = [ (d + n λ/2)2 – d2]1/ = √(n λd) [1+n λ/4d]1/ = √(n λ d) , as we assumed d >>> λ

Let us consider a plane wavefront WW’. The annular area between nth and (n-1)th cricle is known as n th half period zone. The area of the n th half – period zone is given by A n = πr 2 n

  • πr 2 n- = π λ d , Thus the area of all half period zones are approximately equal. The resultant amplitude at point P can be written as u(P) = u 1
  • u 2
  • u 3
  • u 4

n+ u n

The resultant amplitude at P u(P) = u 1 /2, for large n Thus, the resultant amplitude produced by the entire wavefront is only one half of the amplitude produced by the first half-period zone.

Diffraction by a circular aperture

By using the Fresnels half period zones theory we can study the diffraction of a plane wave by a circular aperture. In fig 1. if we increase the radius of circular aperture, when a = √λd, then u (P) = u 1 = 2 ( amplitude for the unobstructed wavefront). and when a = √2λd, then u(P) = (u 1

  • u 2 ) would be almost zero. Thus , by increasing the hole diameter, the intensity at the point P decreases almost to zero. In general, If a = √(2n+1) λd: n = 0, 1, 2, 3… We get maxima.

If instead of circular aperture, we have a circular disc (see Fig.3) and if the disc obstructs the first p half period zones then the resultant field at the point P would be u (P) = u p+

  • u p+

= u p+

Thus, we should always obtain a bright spot on the axis behind a circular disc. This is called the ‘Poisson spot’.

The zone –plate

A zone plate is a beautiful application of the concept of Fresnels half – period zones, which consists of a large number of concentric circles (see Fig4) and have following properties :- (i). The areas of the half period zones are equal. (ii). The redii of half period zones are proportional to the square root of th natural numbers. i.e r n = √nk (iii). If the light is obstructed from alternate zones, the intensity of the image is sufficiently increased. As we know u(P) = u 1

  • u 2 + u 3 –u 4

If the light is obstructed from even half period zones then u(P) = u 1

  • u 3
  • u 5

There are two types of Zone Plate :- Positive and Negative Zone Plate. If the odd (even) numbers are covered with black ink, and an accurate photograph of this is taken on a glass plate of uniform thickness. In the negative of the photograph of the glass plate, the odd (even) zones which are painted black appear transparent and the even (odd) zones appear black. The resulting glass negative is known as positive (negative) zone plate