Question

1. Consider an arrangement of N rectangular slits with a distance d between adjacent slits. The slits emits light rays coherently and in phase at wavelength 4. The electric field at a distant point P is N-1 Ep(r, t) = [Eo cos (kr - wt +no), TO where Eo is the amplitude at P of the electric field due to an individual slit, o = 27d sin 0/X, is the angle of the rays reaching P (as measured from the perpendicular bisector of the slit arrangement) and r, which is much larger than d, is the distance from P to the most distant slit. Show that the electric field at point P is Er(r,t) = E Sin (0)23 cos kr – wt+ (N − 1) , where the quantity in the first square bracket is the amplitude of the electric field at P. [30 pts)

Answer 1

The Electric field at point P is

N-1 Ep(r,t) = Epcos(kr – wt+nd n=0
......................(i) [ where
2ndsino ψ =
]

Let

(kr - wt) = $

So, equation (i) can be written as

N-1 E (r, t) = E.cos(& +no) n=0

  

= E cos()+cos(+0)+cos(+26)+cos(+30)..... +cos($+(N-1)0)

Now Multiplying both sides with we get

2sin(0/2)

Er, t) sin(6/2) % = 2sin(0/2)(cos($)+cos($+$)+cos(+20)+cos(+36) EO
  
...... + cos(+(N-1)

= 2sin(0/2)cos(s) + 2sin(0/2)cos(+6) +2sin(0/2)cos(+20)

  

+2sin(0/2)cos(+30)...... +2sin(0/2)cos(+ (N-1))

Now the first term in the series we can write as

$2 sin( \phi/2)cos(\xi)=sin( \xi+\phi/2)-sin(\xi- \phi/2)$

Similarly, the second term we can write as

$2 sin( \phi/2)cos(\xi+ \phi)=sin( \xi + \phi +\phi/2)-sin(\xi+\phi- \phi/2)$

$=sin( \xi +3\phi/2)-sin(\xi+ \phi/2)$

Similarly for other terms

$2 sin( \phi/2)cos(\xi+ 2\phi)=sin( \xi +2 \phi +\phi/2)-sin(\xi+2\phi- \phi/2)$

$=sin( \xi +5\phi/2)-sin(\xi+ 3\phi/2)$

And

$2 sin( \phi/2)cos(\xi+ 3\phi)=sin( \xi +3 \phi +\phi/2)-sin(\xi+3\phi- \phi/2)$

  $=sin( \xi +7\phi/2)-sin(\xi+ 5\phi/2)$

.................

$2sin( \phi/2)cos(\xi+ (N-1) \phi)]=sin( \xi +(N-1) \phi +\phi/2)-$$sin(\xi+(N-1)\phi- \phi/2)$

$=sin[\xi+(N-1/2)\phi]-sin[\xi+(N-3/2) \phi]$

So,

$2\frac{E_p(r,t)sin( \phi/2)}{E_0}$   $=sin( \xi+\phi/2)-sin(\xi- \phi/2)$$+sin( \xi +3\phi/2)-sin(\xi+ \phi/2)$

$+sin( \xi +5\phi/2)-sin(\xi+ 3\phi/2)$$+sin( \xi +7\phi/2)-sin(\xi+ 5\phi/2)$

$+........+sin[\xi+(N-1/2)\phi]-sin[\xi+(N-3/2) \phi]$

= sin8 + (N - 1/2)] - sinle - 6/2)

  $=2sin (N\phi/2)cos(\frac{2\xi+(N -1)\phi}{2})$

Or

$2\frac{E_p(r,t)sin( \phi/2)}{E_0}=2sin (N\phi/2)cos( \xi +\frac{(N -1)\phi}{2})$

Or

$E_p(r,t)=E_0\frac{sin (N\phi/2)}{sin( \phi/2)}cos( \xi +\frac{(N -1)\phi}{2})$

   Proved.

= E, sin(N6/2), sin (6/2) cos(kr + wt + (N - 1)