Figure 2

Consider the contribution to the field E ⃗ E ⃗ at a P due to a small element of the slit ⅆ y ⅆ y at y y . It is a distance r r from P. R R is the distance from the center of the slit to P.

lets define ε L ε L which is the source strength per unit length, which is a constant.

then d E = ε L r ⅆ y e i ( k r − ω t ) d E = ε L r ⅆ y e i ( k r − ω t )

Now from the drawing r 2 = ( R − y sin θ ) 2 + ( y cos θ ) 2 = R 2 + y 2 sin 2 θ − 2 R y sin θ + y 2 cos 2 θ = R 2 + y 2 − 2 R y sin θ = R 2 [ 1 − 2 y R sin θ + y 2 R 2 ] r 2 = ( R − y sin θ ) 2 + ( y cos θ ) 2 = R 2 + y 2 sin 2 θ − 2 R y sin θ + y 2 cos 2 θ = R 2 + y 2 − 2 R y sin θ = R 2 [ 1 − 2 y R sin θ + y 2 R 2 ] Now assume that y < < R y < < R (which gives us the Franhaufer condition) and r = R [ 1 − 2 y R sin θ ] 1 2 r = R [ 1 − 2 y R sin θ ] 1 2 now expand the square root

r = R [ 1 − y R sin θ + … ] r = R [ 1 − y R sin θ + … ] and neglect higher terms so that r = R − y sin θ r = R − y sin θ thus d E = ε L R e i ( k ( R − y sin θ ) − ω t ) ⅆ y d E = ε L R e i ( k ( R − y sin θ ) − ω t ) ⅆ y where now we have used R in the denominator since it is much bigger than y d E = ε L R e i ( k R − ω t ) e − i k y sin θ ⅆ y d E = ε L R e i ( k R − ω t ) e − i k y sin θ ⅆ y

now integrate assuming that θ θ is a constant over the slit E = ε L R e i ( k R − ω t ) ∫ − a / 2 a / 2 e − i k y sin θ ⅆ y = ε L R e i ( k R − ω t ) e − i k y sin θ − i k sin θ | − a / 2 a / 2 = ε L R e i ( k R − ω t ) e − i k a 2 sin θ − e i k a 2 sin θ − i k sin θ = ε L R e i ( k R − ω t ) − 2 i sin ( k a 2 sin θ ) − i k sin θ = ε L R e i ( k R − ω t ) 2 sin ( k a 2 sin θ ) k sin θ = ε L a R e i ( k R − ω t ) sin ( k a 2 sin θ ) k a 2 sin θ E = ε L R e i ( k R − ω t ) ∫ − a / 2 a / 2 e − i k y sin θ ⅆ y = ε L R e i ( k R − ω t ) e − i k y sin θ − i k sin θ | − a / 2 a / 2 = ε L R e i ( k R − ω t ) e − i k a 2 sin θ − e i k a 2 sin θ − i k sin θ = ε L R e i ( k R − ω t ) − 2 i sin ( k a 2 sin θ ) − i k sin θ = ε L R e i ( k R − ω t ) 2 sin ( k a 2 sin θ ) k sin θ = ε L a R e i ( k R − ω t ) sin ( k a 2 sin θ ) k a 2 sin θ

now we define β = k a 2 sin θ β = k a 2 sin θ and see that we can rewrite our expression as E = ε L a R sin β β e i ( k R − ω t ) E = ε L a R sin β β e i ( k R − ω t ) or equivalently E = ε L a R s i n c β e i ( k R − ω t ) E = ε L a R s i n c β e i ( k R − ω t )

The intensity will go like the square of this so

I = I 0 s i n c 2 β I = I 0 s i n c 2 β

Figure 3

The Intensity has a maximum at β = 0 β = 0 or θ = 0 θ = 0 . there are minima when sin β = 0 sin β = 0 or β = k a 2 sin θ = n π β = k a 2 sin θ = n π 2 π λ a 2 sin θ = n π 2 π λ a 2 sin θ = n π sin θ = n λ a sin θ = n λ a in the case of small θ θ we see that Δ θ = λ a Δ θ = λ a is the distance between adjacent minima.

As a a becomes large, we see that the minima will merge together. This is consistent with what we said at the beginning, that if a > > λ a > > λ then you just get shadowing but not diffraction.

Finding the secondary maxima is more difficult. (Take the derivative of I and then look for zeros.) This can not be done analytically.

Note that wee have been considering only one dimension. If the length of the slit is L L then we have only considered the case that L > > λ L > > λ and so diffraction occurs only in the other dimension.