DiffractionLecture 1Dr. Aparna TripathiWhat is diffraction?
Diffraction is bending of waves around obstacles. Diffraction also
occurs when waves from a large number of sources interfere.Dr.
Aparna TripathiDiffractionServes as a pointto generate newwavesNew
Wave FrontsDr. Aparna TripathiThe narrower the slit, the more the
slit behaves like a point sourceThe wider the slit, the more like a
plane wave the light is A plane wave does not bend at the slit if
the openingd >>.If > d, the bending takes placeDr. Aparna
TripathiHuygenss PrincipleDr. Aparna TripathiDiffraction dudes:
Fresnel and FraunhoferDr. Aparna TripathiFraunhofer: Source and
screen are at infinite distance from slit. Plan wavefrontConverging
deviceFresnel: Source and screen are at finite distance from
slit.Spherical wavefrontNo converging devicePSFresnel vs.
Fraunhofer diffractionDr. Aparna
TripathibA1A2A3AnA2ALfPBnB2B3B1uCFraunhofer diffraction pattern
produced by an infinitely long slit of width b A plane wave is
incident normal on the slit. According to Huygens theory each point
on slit spends out secondary wavelets in all direction. All the
wavelets start from various points in slit in the same phase. The
ray diffracted along the direction of incident rays are focused at
C. The point C is optically equidistant from all the points on the
slit. Therefore all the secondary wavelets from slit reach to C in
the same phase.Hence there is maximum intensity at C.The rays
diffracted through angle u are focused at P.Let us find out the
resultant intensity at P. Let the point sources be at A1, A2,
A3,An.The distance between two consecutive points be A.If the
number of point sources be n thenb = (n-1)ASingle Slit Diffraction
Pattern Let us draw A1A2 perpendicular to A2B2 The optical path
A1B1P and A2B2P are sameThe path differencewould be A2A2 = A sin
uDr. Aparna Tripathi If the path difference is equal to the
wavelength of the light used, then P will be a point of minimum
intensity u n bsin = The path difference is odd multiples of /2 the
direction of secondary maxima ( )21 n 2bsinu+= The diffraction
pattern due to single slit consist of a central bright maxima at C
followed by secondary maxima and minima at both the sideDr. Aparna
TripathiIntensity distribution in diffraction pattern due to single
slit bA1A2A3AnA2ALfPBnB2B3B1uCThe path differencewould be A2A2 = A
sin uDr. Aparna TripathiThe corresponding phase difference would be
....2 sin2ut| A - =If the field at the point P due to the
disturbance emanating from point A1A1= a cos(et)field at the point
P due to the disturbance emanating from point A2A2= a cos(et-|)A3=
a cos(et-2|)...An= a cos[et-(n-1)|]Then the resultant field at the
point PE= A1+A2+A3..............+AnDr. Aparna TripathiE=a cos(et) +
a cos(et-|) + a cos(et-2|) + + a cos[et-(n-1)|]( ) | | ( ) | | |
e||| e | e | e e 121cos2sin2sin1 cos ) 2 cos( ) cos( ) cos( = + . .
. . . + + + n tnn t t t t( ) | |( ) | | 3 ....... 121cos E
EE2sin2sinfield resultantthe of amplitude the
where121cos2sin2sinEthus| e||| e||uu == =n tnan tnaDr. Aparna
Tripathi4 ......sinsinsinsin22sinEsin 2sin2Furthersin2sin2have web
n0 and n limitIn the|| u t u t||utut|ut |ut|uAbbnananbbnn n=
=|.|
\|= - = A - =- =A - = A A Dr. Aparna Tripathi( )...8 sin sinIby
givenis ondistributi intensitying Correspond...7
cossin...6bsin...5220222| || || e||I At A Ethusna Awhere= = ===Dr.
Aparna TripathiCentral Maximum:For the pt C on the screen u= 0;
hence |=0.| 0 value of sin | / | = 1Hence the intensity at P
020sinI I I =||.|
\|=||Secondary Maximum:The directions of secondary maxima are
given by( )( ) ( )etcn nbnn27,25,23eq above in1,2,3.... nPutting2 1
2 2b1 2 b6 eq inof value the putting21 2sinn==+=+=+=uuDr. Aparna
TripathiFor first secondary maximum:23 =22 942323sinsinI0202020I II
I = =||||.|
\|=||.|
\|=ttt||For secondary maximum:25 =intensity decreasing of are
maxima seondarythe thus61 2542525sinsinI0202020I II I = =||||.|
\|=||.|
\|=ttt||Dr. Aparna Tripathi( )( )| || | | | ||||| | | | ||||
|||== =((
=||.|
\|||.|
\|=||.|
\|=tan0 2 sin cos sin 20 maximum I for 2 sin cos sin 2sinsinIby
givenis P pointatIntensity 2 242 202020ddId I dIIddddIISolve the
above eq graphically by plotting the curves y=| and y=tan |Dr.
Aparna TripathiSecondary Minima:The directions of secondary minima
are given byetcnn bn3 , 2 , eq above in1,2,3.... nPutting n6 eq
inof value the puttingsinnt t ttu u=== ==0sinIof values for the
intensity 20=||.|
\|=|||IDr. Aparna TripathiDr. Aparna Tripathi
