Concept of DIFFRACTION-PHYSICS-1, Exams of Physics

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Physics - I

15B11PH

DIFFRACTION

DIFFRACTION

Diffraction of the light occurs when a light wave passes very close to the edge of an object or through a tiny opening, such as a slit or aperture.

LECTURE 8

If observation are made carefully and width of the slit is not

very large with respect to wavelength then the light intensity

in the region AB is not uniform and there is also some

intensity inside the geometrical shadow.

This spreading out of a wave when it passes through a

narrow opening is known as diffraction pattern.

Diffraction is a wave effect

Interference pattern of light and dark bands around the edge of the object.

Diffraction is often explained in terms of the Huygens principle, which states that each point on a wave front can be considered as a source of a new wave.

All points on a wave front serve as point sources of spherical secondary wavelets. After a time t, the new position of the wave front will be that of a surface tangent to these secondary wave fronts

FROUNHOFFER

• In “Fraunhofer” class

of diffraction, incident

& outgoing rays are

parallel.

• Start with single plane

of atoms (a grating)

• Wave “in phase” –

incident crests and troughs aligned

COMPARISON

• In Fraunhofer class, „S‟ and „P‟ are at infinite distance from the aperture and hence both incident and diffracted waves may be considered to be plane wave.
• If source „S‟ and screen „P‟

are at finite distance from the diffraction aperture (or slit system), then we have Fresnel class of diffraction

Fresnel Fraunhofer

Resolving the amplitude parallel and perpendicular to OA

...............( 2 )

sin sin sin 2 sin 3 ... sin( 1 )

................( 1 )

cos cos cos 2 cos 3 ... cos( 1 )

    

    

     

      

R a a a a n

R a a a a a n

Multiply eq. (1) with 2sinφ/

2

... 2 cos( 1 ) sin 2

2 cos 3 sin 2

2 cos 2 sin

2

2 cos sin 2

2 sin 2

2 cos sin

 

 

 

 

  

   

  

a a a n

R a a

sin  A  B  sin A  B   2 cos A sin B

sin[( 1 )

) ... sin[( 1 ) 2

sin( 2

) sin( 2 2

) sin( 2

sin( 2

2 sin

2 cos sin

n

a n

R

sin 2

sin 2

2 R cos sin a a n

  .......... ....( 4 ) 2

sin 1

2

sin

2

sin sin

 



  n 

n a R

Squaring and adding equation (3) and (4)

.......... ....( 5 )

2

sin

2

sin

2

2 2 2 

n  a R 

  ............( 3 ) 2

1 cos

2

sin

2

sin cos

 





 

n

n a R

Similarly

Dividing (4) by (3) we have

tan tan

  

 

n n

n

If we have large number of vibration as

,cos 1

cos ,cos ,

... ωt n

ωt ωt

If we simplify it (Ghatak, superposition of waves)

E 

 

   

 

  

   

 

  

   

 

  



  

   

 

  







( 1 ) 2

1 cos

( 1 ) 2

1 cos

2

sin

2

sin

cos 1

cos cos

R t n

t n

n

... ωt n

ωt ωt E a

Where amplitude R of the resultant wave will be

that n b

in thelimit of n and 0 in such a way

2

sin

2

sin

 

   

 

  



 

n 

R a





 



  sin sin

2

n b

n   