Diffraction_Bsc

aDiffraction of LightDiffraction of Light : The phenomenon of bending of light waves around edges of small obstacles and hence it's spreading into the geometrical shadow of the obstacle is called diffraction. The diffraction effects were first observed by Grimaldi in 1665. The effects can be observed only when the size of the obstacle is very small and comparable to the wavelength of light.

This phenomenon shows that the rectilinear propagation of light (light traveling along a straight) is only approximate i.e. light bends at the corners of small obstacles and enters the regions of geometrical shadows.Light from the source S is made to fall on a slit AB whose width is very small. The region CD on the screen is found to contain unequally spaced alternate bright and dark fringes with light bending into the region above C and below D. This is due to diffraction effects. Fresnel explained this phenomenon by applying the Huygens ‘Principle along with the principle of interference. Diffraction phenomenon is classified into two types

Fresnel's diffraction:

Fraunhofer's diffraction

1. The source of light and the screen on1. which the diffraction pattern is observed are at finite distance from the obstacle or aperture.2. The incident wavefront and thediffracted wavefronts are spherical or2. cylindrical.3. The incident beam is a divergent3. beam whereas the diffracted beam is3. a convergent beam.4. No changes in the wavefront are4. made by using either lenses or4. mirrors.5. The centre of the diffraction pattern is5. either bright or dark. The pattern is5. the image of the obstacle or aperture.

1. The source of light and the screen on1. which the diffraction pattern is observed are at infinite distance from the obstacle or aperture.2. The incident wavefront and the2. diffracted wave fronts are plane2. wave fronts.3. The incident beam is a parallel beam3. and the diffracted beam is also3. parallel beam.4. The incident rays from a source are4. made parallel using a convex lens4. and the diffracted rays are brought 4. to focus on a screen using another4. convex lens (converging lenses).5. The centre of the diffraction pattern5. is always bright. The pattern is the5. image of the source itself.

Examples of diffraction – (1) The luminous border that surrounds the profile of amountain just before sun rises behind it, (2) the light streaks that one sees whilelooking at a strong source of light with half shut eyes and (3) the coloured spectraone sees while viewing a distant source of light through a fine piece of cloth.Fresnel’s assumptionsFresnel in 1815, combined the Huygens principle of wavelet and the principle ofinterference to explain the bending of light around obstacles and also the rectilinearpropagation of light.1. According to Huygens’ principle, each point of a wavefront (wavefront is a locus of points in a medium that are vibrating in same phase) is a source of secondary disturbance and wavelets coming from these points spread out in all directions with the speed of light. The envelope of these waves constitute the next wavelet.2. According to Fresnel, a wavefront can be divided into a large number of strips or zones called Fresnel zones of small area. The resultant effect at any point will depend on the combined effect of all the secondary waves coming from various zones.3. The effect at a point due to any particular zone depends on distance of the point from the zone.4. The effect will also depend on the obliquity (inclination) of the point with reference to the zone under consideration. Division of wavefront into Fresnel’s half period zones – Expressionfor resultant displacement/amplitude – Rectilinear propagation of lightABCD is a plane wave front of monochromatic light of wavelength 𝜆. The diagram shows the plane wavefront as perpendicular to plane of the paper. Consider a point P at a distance b from the wave front at which amplitude due to the wave is to be found.

To find the resultant amplitude at P due to entire wavefront, Fresnel assumed thewavefront to be divided into a number of concentric half period zones.Fresnel’s half period zones: With P as centre and with M1P=(b+𝜆

2),M2P=(b+2𝜆

2),….. as radii, a series of concentric spheres are drawn on the wavefront. These spheres intersect the wavefront in concentric circles. These circles or zones are of radii OM1,OM2,…. on the wavefront with O as centre.The secondary waves from any two consecutive zones reach the point P with a path difference of 𝜆

2 or a time period of T

2. Hence these zones are called half period zones. The area of the circle OM1 is called first half period zone. The area between the circles of OM2 and OM1 is called second half period zone and so on. The area between the nth and (n-1)th circle is called the nth half period zone. To find the radius of a half period zone:In the diagram, from the right angled triangle OM1P,a

	OM1=(M1P)2-(OP)2=(b+𝜆 2)2-b2
	OM1=(b2+2b𝜆 2+𝜆2 4)-b2 or OM1=b𝜆 ( neglecting 𝜆2 4 as b≫𝜆)

OM1=b𝜆 is the radius of first half period zone.The radius of the second half period zone isOM2=(M2P)2-(OP)2=(b+2𝜆

2)2-b2 Thus OM2=2b𝜆 Similarly the radius of the nth half period zone is OMn=(b+n𝜆

2)2-b2 or OMn=nb𝜆Thus the radii of 1st ,2nd ,……. half period zones are b𝜆,2b𝜆,……nb𝜆.Therefore, the radii of the zones are proportional to the square root of natural numbers. To find the area of half period zones :The area of first half period zone is a

	=𝜋(OM1)2=𝜋[(M1P)2-(OP)2]( As area =𝜋r2)
	=𝜋[(b+𝜆 2)2-b2]=𝜋b𝜆

The area of 2th half period zone =𝜋[(OM2)2-(OM1)2]=𝜋[2b𝜆-b𝜆]=𝜋b𝜆 The area of nth half period zone =𝜋[(OMn)2-(OM1)2]=𝜋[nb𝜆-b𝜆]=𝜋b𝜆 Thus the area of each half period zone is same and is equal to 𝜆.Also, the area of any zone is directly proportional to the wavelength ( 𝜆 ) of light and the distance of the point from the wavefront (b). To find amplitude due to the wavefront:The amplitude of the waves at P due to an individual zone is1. Directly proportional to the area of the zone2. inversely proportional to the distance of the point P from the given zone.3. the obliquity factor (1+cos𝜃) where 𝜃 is the angle between normal to the zone and the line joining the zone to the point P . The effect at P decreases as obliquity increases.The path difference between any two consecutive half period zones is 𝜆

2. Hence the waves from two consecutive zones will reach P in opposite phase. If m1,~m2,~m3,….. are the amplitudes at P due to 1st ,2nd ,3rd ,….. half period zones, the resultant amplitude at P due to entire wavefront isA=m1-m2+m3-m4+⋯.+mn if n is oddand A=m1-m2+m3-m4+⋯.-mn if n is even.As the obliquity increases amplitudes decreases, ie. m2 is less than m1,m3 is less than m2 etc...Thus on the average m2=m1+m3

2… (1) Similarly m4=m3+m5

2.The equation A=m1-m2+m3-m4+⋯. can be written asA=m1

2+(m1

2-m2+m3

2)+(m3

2-m4+m5

2)+⋯ Substituting equations (1) and (2) in (3) we getA=m1

2+mn

2 if n is odd ....(5) (The terms in the bracket cancel)A=m1

2+mn-1

2-mn if n is even. ....(6).As the amplitudes are of diminishing order, for large n,mn and mn-1 tend to zero. Thus A=m1

2.The amplitude of the wave at any point P , in front of a large plane wavefront is equal to half the amplitude due to the first half period zone.As the intensity is proportional to square of the amplitude, (I∝A2| the intensity at P is proportional to m12

4(I∝m12

4). Thus the intensity at point P is one fourth of the intensity due to the first half period zone. Explanation of rectilinear propagation of lightThe intensity at point in front of a wave front is proportional to m12

4 where m1 is the amplitude of the first half period zone. Thus the intensity at point P is one fourth of the intensity due to the first half period zone.Thus only half the area of the first half period zone is effective in producing the illumination at the point P. A small obstacle of the size of half the size of half the area of first half period zone placed at O will block the effect of whole wavefront and the intensity at P due to rest of the wavefront is zero.While dealing with the rectilinear propagation of light, the size of the obstacle used is far greater than the area of the first half period zone and hence the bending effect of light round the corners of the obstacle is diffraction effects cannot be noticed.Thus if the size of the obstacle placed in the path of light is very small and comparable to wavelength of light, then it is possible to observe illumination in the region of the geometrical shadow also. Thus, rectilinear propagation of light is only approximately true.Zone plateA zone plate is a specially constructed screen such that light is obstructed from everyalternate zone.The correctness of Fresnel’s method in dividing a wavefront into halfperiod zones can be verified with the help of zone plate. A zone plate is constructed by drawing concentric circles on a white paper such that radii are proportional to the square root of the natural numbers. The odd numbered zones (i.e. 1st, 3rd, 5th …) are covered with black ink and a reduced photograph is taken. The negative of the photograph appears is as shown in Fig. (a). The negative shows odd zones are transparent to incident light and even zones will cut off light. This is a positive zone plate. If odd zones are opaque and the even zones are transparent then it is a negative zone plate. Fig. (b)

TheoryLet S be a point source of light of wavelength 𝜆 placed at a distance a from centre O of the zoneplate. Let P be the point on a screen placed at distance b at which intensity of diffracted lightbright.

Let r1,r2,r3…….rn be the radii of the 1st ,2nd ,3rd …….nth half period zones respectively. The position of the screen is such, that from one zone to the next there is an increasing path difference of 𝜆

2.Thus, from the diagram SO+OP=a+bSM1+M1P=a+b+𝜆

2 Similarly SM2+M2P=a+b+2𝜆

2 and so onFrom the triangle SM1O SM1=(SO2+OM21)1/2=(a2+r21)1/2Similarly from the triangle PM1O M1P=(OP2+OM21)1/2=(b2+r21)1/2Substituting the values of SM1 and M1P in equation (1), we geta

	(a2+r21)1/2+(b2+r21)1/2=a+b+𝜆 2
	or a(1+r21 a2)1/2+b(1+r21 b2)1/2=a+b+𝜆 2

Expanding and simplifying the above equation, we geta

	a(1+r21 2a2)+b(1+r21 2b2)=a+b+𝜆 2
	a+r21 2a+b+r21 2b2=a+b+𝜆 2
	or r21 2(1 a+1 b)=𝜆 2 or r21(1 a+1 b)=𝜆

Thus for the radius of the nth zone the above relation can be written asr2n(1

a+1

b)=n𝜆…..(2) or r2n=ab

a+bn𝜆 or rn=ab𝜆

a+bn Thus the radii of the half period zones are proportional to the square root of the natural numbers.From equation (2) can written as (1

a+1

b)=n𝜆

r2nThis equation is similar to the lens formula 1

u+1

v=1

fComparing equations (3) and (4) 1

f=n𝜆

r2n or f=r2n

n𝜆f is the focal length of zone plate and acts as a convex lens of multiple foci.The path difference between any successive transparent zones is 𝜆 and the phase difference is 2𝜋. Waves from successive zones reach P in phase. Focussing action of Zone plateThe amplitude at P depends on (a) area of the zone, (b) distance of the zone from P and (c) obliquity factor.The area of nth zone =𝜋r2n-𝜋r2n-1As r2n=ab

a+bn𝜆, the area of the nth zone =ab

a+bn𝜆-𝜋ab

a+b(n-1)𝜆=𝜋ab𝜆

a+bArea is independent of n . Area of all zones are same. But the distance of the zone from P and obliquity factor increases as n increases.The resultant amplitude at P isA=m1+m3+m5+… for positive zone plateA=-(m2+m4+m6+……) for negative zone plate.This is much greater than A=m1

2 which is due to all zones.As the intensity from the zone plate is very high, the zone plate is said to have focussing action.

Diffraction at a straight edgeLet a cylindrical wavefront from a rectangular slit ‘S’, illuminated by amonochromatic source of wavelength ‘’, falls on a sharp and straight edge of an opaqueobstacle AB [such as blade] such that the edge of the blade is parallel to the silt. Join SA andproduce it to meet the screen CD placed perpendicular to the plane of the paper at ‘O’. According to geometric optics, the regionbelow ‘O’ should have shadow of the obstacle andthe region above ‘O’ should be illuminated. But, experiments show that the intensity falls offcontinuously and rapidly as we move deep into theregion below ‘O’ and becomes totally dark after ashort distance. Just above ‘O’, alternate dark and bright bandsof unequal width and varying intensity parallel tothe straight edge are observed. The bands becomecloser and less distinct as we move away from ‘O’until the screen is uniformly illuminated.