POWER POINT PRESENTATION

TOPIC : FRESNEL'S THEORY OF

DIFFRACTION

BY:- PROF.S.V.ANGADI, PHYSICS DEPARTMENT, J.T.COLLEGE, GADAG

FRESNEL'S THEORY OF DIFFRACTION

FRESNEL'S EXPLANATION

• The diffraction phenomenon is caused by the interference of innumerable secondary wavelets produced by the unobstructed portions of the same wave-front.

FRESNEL'S ASSUMPTIONS:

i) Each element of a wave-front sends secondary waves continuously.

ii) A wave front can be divided into a large number of strips or zones called the Fresnel's zones

ii)The resultant effect at any external point P is determined by combining the effects of all the secondary waves reaching them from various zones.

iv) The effect at a point due to any particular zone depends on the distance of the point from the zone.

v) The effect at any point P depends on the inclination of the point with reference to zone under consideration. The intensity is maximum along a direction normal to the zone and decreases as the angle of inclination Increases.

FRESNEL'S HALF PERIOD FOR A PLANE WAVE :

• Let ABCD be a section of a plane wave-front of monochromatic light of wave length , traveling from left to right (fig). Let P be an external point at which the effect of the entire wave-front is desired.

Let OP is perpendicular drawn from P to the wave-front and is equal to b. To find the resultant intensity at P due to the wave-front, by Fresnel’s method, the wave-front is divided In to a number of concentric half period zones called “Fresnel's zone." and then the effect of all the zones at point p is found. This can be done as follows: Considering P as centre and radii equal to b+𝜆/2, b+2𝜆/2, b+3𝜆/2,... etc. draw a series of spheres on the wave-front thus cutting the wave-front into annular strips or zones. The sections of these spheres by the plane wave-front are concentric circles having common centre O and radii 0𝑀1, 0𝑀2, 0𝑀3...0𝑀𝑛−1, 0𝑀𝑛 ......etc.

The secondary wavelets from any two consecutive zones reach P with a path difference 𝜆/2 or time difference half period. That is why the zones are called half period zones. THE area of the innermost (first) circle is called first half period zone; similarly the annular area between the first circle and the second circle is called second half period zone and so on. Thus the annular area between (n-1)th and nth circle is called nth half period zone. The point O is called the pole of the wave-front with respect to point P. A Fresnel half period zone with respect to an external point P is a thin annular zone or a thin strip of the primary wave-front surrounding the point O such that the distances of its outer and inner edges from O differs by 𝜆/2.

RADII OF HALF PERIOD ZONE:

The radius of first half period zone, OM1 = ඥሺM1Pሻ2 −ሺOPሻ2

OM1 = ඨ൬b+ λ2൰2 − b2 OM1 = ඨb2 + λ42 + bλ− b2 OM1 = ξbλ ,As b >> 𝜆

The radius of second half period zone,

OM2 = ඥሺM2Pሻ2 −ሺOPሻ2

OM2 = ඨ൬b+ 2λ2൰2 − b2 OM2 = ඨb2 + 4λ4 2 + 2bλ− b2

OM2 = ξ2bλ

Similarly the radius of the nth half period zone is,

OMn = ඥሺMnPሻ2 −ሺOPሻ2

OMn =ඨ൬b+ nλ2൰2 − b2 OMn =ඨb2 + (nλ)4 2 + nbλ− b2

OMn = ξnbλ Thus we see that the radii of half period zones are proportional to the square root of the natural numbers.

Area of a half period zone:

As area of nth zone is independent of n, thus the area of each half period zone is approximately the same and is equal More accurately the area of the zone increases slightly with n.

The area of n zone= πሾሺOMnሻ2 −ሺOMn−1ሻ2ሿ = πሾሺPMnሻ2 −ሺPOሻ2ሿ− πሾሺPMn−1ሻ2 −ሺPOሻ2ሿ

= πቀb+ nλ2ቁ2 −ሺbሻ2൨− πቀb+ (n−1)λ2 ቁ2 −ሺbሻ2൨

= πቂb2 + ሾnλሿ24 + bnλ− b2ቃ− πቂb2 + ሾ(n−1)λሿ24 + bሾn− 1ሿλ− b2ቃ = πቈሾnλሿ24 + bnλ − πቈሾ(n− 1)λሿ24 + bሾn− 1ሿλ

= πቈnλ+ λ24 ሺ2n− 1ሻ = πbλ

THE DISTANCE OF THE POINT FROM THE HALF PERIOD ZONE:

The average distance of nthhalf period zone from point P,

= ቂb+ nλ2ቃ+ +b+ (n− 1) λ22 = b+ (2n− 1) λ4

THE AMPLITUDE OF THE DISTURBANCE AT P DUE

TO AN INDIVIDUAL ZONE :The amplitude of the disturbance due to a given zone is (i) Directly proportional to the area of the zone because number of point sources, from each of which a secondary wavelet starts, in a zone are proportional to the area, (ii) Inversely proportional to the distance of the point P from the given zone. (iii) Directly proportional to the obliquity factor (1+cos 𝜃) whose 𝜃is the angle between the normal to the zone and the line joining the zone to point P.

Thus amplitude of the disturbance at P due to 𝑛𝑡ℎ zone is,

𝑅𝑛𝛼𝜋𝑏𝜆+ 𝜆2ሺ2𝑛− 1ሻ4 ൨

𝑏+ሺ2𝑛− 1ሻ𝜆4 ሺ1− 𝑐𝑜𝑠𝜃𝑛ሻ 𝑅𝑛𝛼 𝜋𝜆ሺ1− 𝑐𝑜𝑠𝜃𝑛ሻ As n increases, 𝜃𝑛 increases and cos 𝜃𝑛 decreases. Thus the amplitude of the disturbance at P due to a given zone decreases as the order of the zone increases. This means that the amplitude of the disturbance due to first half period zone is maximum and it decreases regularly as we pass from the inner zone to the next outer.

THE RESULTANT AMPLITUDE DUE TO THE WHOLE WAVE-FRONT:

Let 𝑅1,𝑅2, 𝑅3, …..𝑅𝑛 be the amplitudes of the disturbances at P, due to the first second, third...., nth half period zones respectively. The magnitudes of 𝑅1,𝑅2 etc are of continuously in decreasing order. As the path difference between the wave reaching P from any two consecutive half-period zones is 𝜆/2, the waves from two consecutive zones reach P in the opposite phase.

Therefore if amplitude due to first zone is positive, that due to second zone is negative, that due to third zone is positive and so on, i e 𝑅1,𝑅3, …..𝑅𝑛−1 etc are positive and 𝑅2, 𝑅4, …..𝑅𝑛 etc, are negative Hence the resultant amplitude at P due to the entire wave front is 𝑅= 𝑅1 − 𝑅2 + 𝑅3,….(−1)n−1𝑅𝑛 As the magnitudes of successive terms 𝑅1,𝑅2, 𝑅3, …..𝑅𝑛 decrease gradually, 𝑅2slightly less than 𝑅1 and greater than 𝑅3 so that we may write, 𝑅2 = 𝑅1 + 𝑅32

𝑅3 = 𝑅2 + 𝑅52

And so on.

Now equation may be written in the form 𝑅= 𝑅12 +ቀ

𝑅12 − 𝑅2 + 𝑅32ቁ+ቀ𝑅32 − 𝑅+ 𝑅52ቁ+………𝑅𝑛2

if n is odd. And 𝑅= 𝑅12 +ቀ

𝑅12 − 𝑅2 + 𝑅32ቁ+ቀ𝑅32 − 𝑅3 + 𝑅52ቁ+………𝑅𝑛−12 –𝑅𝑛

if n is even.

In above relations the quantities in the bracket is very nearly equal to zero hence we can write 𝑅= 𝑅12 + 𝑅𝑛 2 𝑖𝑓 𝑛 𝑖𝑠 𝑜𝑑𝑑

𝑅= 𝑅12 + 𝑅𝑛 2 𝑖𝑓 𝑛 𝑖𝑠 𝑒𝑣𝑒𝑛

But usually n is large, hence we may write 𝑅𝑛−1 (appro).then 𝑅𝑛−1 2 − 𝑅𝑛 = −𝑅𝑛 2

So that above two equations may be represented by a single equation as 𝑅= 𝑅12 ± 𝑅𝑛 2

The plus and minus sign being taken accordingly as n is odd or even. For large wave front, n is very large, hence Rn vanishes; so that we can have 𝑅= 𝑅12

Thus the amplitude due to a large wave front at a point in front of it is half that due to the first half period zone acting alone. The intensity at any point is proportional to the square of the amplitude, therefore the resultant Intensity at P 𝐼𝛼𝑅𝑛4

This means the intensity at appoint is one forth of intensity due to the first half period zone alone.

EXPLANATION OF RECTILINEAR PROPAGATION OF LIGHT:

Let a plane wave-front of monochromatic light be incident normally on a rectangular aperture ABCD shown in figure. Let a screen be placed parallel to the rectangular aperture at some distance from it. Let A'B'C'D' (full line rectangle on the screen) be the geometrical projection of ABCD on the screen. If the law of rectilinear propagation were strictly true, we would obtain uniform illumination inside the full line rectangle and uniform darkness outside it.

Let us consider four typical points P1, P2, P3, P4 on the screen. The corresponding poles are O1 O2, O3, &O4 respectively. The point P1 is well inside the full rectangle. Its pole O1 is at large distance from the edge of the aperture as compared with the wavelength of light. Therefore if we draw half period zones with O1, as centre, a large number of zones can be drawn before the edge of the aperture is reached. Thus all the effective zones are open and hence the resultant amplitude at is equal to half the amplitude due to first half period zone. As the area of these half period zones is extremely small, we can conclude that light from O1 has traveled to P1 along a straight line. The points in the neighborhood of P1 are similarly and uniformly illuminated.

The point P2 is well inside the geometrical shadow. Its pole is O2. If we draw half period zones around O2, all the effective zones are totally blocked and hence the resultant amplitude at P2 is zero. Thus there is complete darkness at P2.The point P3 Is near the edge and inside the geometrical projection. Its pole 01,is near the edge AD of the aperture. If half period zones are drawn around O3, they are partly cut off and the areas of the zones change too rapidly so that the above theory cannot be applied. Similarly for point P4 which is near the edge and outside the geometrical projection, the intensity cannot be decided.

Thus there is uniform illumination at all points inside the inner dotted rectangle and complete darkness at all points outside the outer dotted rectangle. P1 points lying between the two dotted rectangles, this elementary theory gives no information. It is certain that at these points here is neither nether illumination nor complete darkness. As the wavelength of light is very small, the two dotted rectangles lie very close to the full line rectangle. Thus Fresnel’s wave theory accounts for approximate rectilinear propagation of light

THE END