0.1 Single slit diffraction

Diffraction

Diffraction is an important characteristic of waves. It can be said to one of the defining characteristics of a wave. It occurs when part of a wavefront isobstructed. The parts of the wavefronts that propagate past the the obstacle interfere and create a diffraction pattern. Diffraction and interference areessentially the same physical process, resulting from the vector addition of fields from different sources. By convention interference refers to only a fewsources and diffraction refers to many sources or a continuous source.

When a plane wave hits an aperture, Huygens principle says that each point in the aperture acts as a source of spherical wavelets. The maximum path lengthdifference of all these sources is between the top and the bottom. $$\Delta r_{max}=a\sin \theta$$ The waves start out in phase. If $a<<\lambda$ then the slit acts as a point source and you get a spherical wave coming out. If $a>>\lambda$ then the aperture simply casts a bright spot the size of the aperture shadow. But if $\lambda \approx a$ then an interference pattern is set up.When the resulting pattern is viewed close to the aperture, the pattern can be very complex, and this is callFresnel diffraction. As the the pattern is viewed from further and further away, it eventually stops changing shape and only grows in size. This isFraunhoffer diffraction.

Single slit diffraction

Consider the contribution to the field $\vec{E}$ at a P due to a small element of the slit $dy$ at $y$ . It is a distance $r$ from P. $R$ is the distance from the center of the slit to P.

lets define ${\epsilon }_L$ which is the source strength per unit length, which is a constant.

then $$dE=\frac{{\epsilon }_L}{r}dy{e}^{i(kr-\omega t)}$$

Now from the drawing $${\displaystyle \begin{array}{c}{r}^2={(R-y\sin \theta )}^2+{(y\cos \theta )}^2\\ =R^2+y^2{\sin }^2\theta -2Ry\sin \theta +y^2{\cos }^2\theta \\ =R^2+y^2-2Ry\sin \theta \\ =R^2\left[1-\frac{2y}{R}\sin \theta +\frac{y^2}{R^2}\right]\end{array}}$$ Now assume that $y<<R$ (which gives us the Franhaufer condition) and $$r=R{\left[1-\frac{2y}{R}\sin \theta \right]}^{\frac{1}{2}}$$ now expand the square root

$$r=R\left[1-\frac{y}{R}\sin \theta +\dots \right]$$ and neglect higher terms so that $$r=R-y\sin \theta$$ thus $$dE=\frac{{\epsilon }_L}{R}{e}^{i(k(R-y\sin \theta )-\omega t)}dy$$ where now we have used R in the denominator since it is much bigger than y $$dE=\frac{{\epsilon }_L}{R}{e}^{i(kR-\omega t)}{e}^{-iky\sin \theta }dy$$

now integrate assuming that $\theta$ is a constant over the slit $${\displaystyle \begin{array}{c}E=\frac{{\epsilon }_L}{R}{e}^{i(kR-\omega t)}{\int }_{-a/2}^{a/2}{e}^{-iky\sin \theta }dy\\ =\frac{{\epsilon }_L}{R}{e}^{i(kR-\omega t)}\frac{e^{-iky\sin \theta }}{-ik\sin \theta }{|}_{-a/2}^{a/2}\\ =\frac{{\epsilon }_L}{R}{e}^{i(kR-\omega t)}\frac{e^{-i\frac{ka}{2}\sin \theta }-e^{i\frac{ka}{2}\sin \theta }}{-ik\sin \theta }\\ =\frac{{\epsilon }_L}{R}{e}^{i(kR-\omega t)}\frac{-2i\sin (\frac{ka}{2}\sin \theta )}{-ik\sin \theta }\\ =\frac{{\epsilon }_L}{R}{e}^{i(kR-\omega t)}\frac{2\sin (\frac{ka}{2}\sin \theta )}{k\sin \theta }\\ =\frac{{\epsilon }_{L}a}{R}{e}^{i(kR-\omega t)}\frac{\sin (\frac{ka}{2}\sin \theta )}{\frac{ka}{2}\sin \theta }\end{array}}$$

now we define $$\beta =\frac{ka}{2}\sin \theta$$ and see that we can rewrite our expression as $$E=\frac{{\epsilon }_{L}a}{R}\frac{\sin \beta }{\beta }{e}^{i(kR-\omega t)}$$ or equivalently $$E=\frac{{\epsilon }_{L}a}{R}sinc\beta e^{i(kR-\omega t)}$$

The intensity will go like the square of this so

$$I=I_{0}sin{c}^2\beta$$

The Intensity has a maximum at $\beta =0$ or $\theta =0$ . there are minima when $\sin \beta =0$ or $$\beta =\frac{ka}{2}\sin \theta =n\pi$$ $$\frac{2\pi }{\lambda }\frac{a}{2}\sin \theta =n\pi$$ $$\sin \theta =\frac{n\lambda }{a}$$ in the case of small $\theta$ we see that $$\Delta \theta =\frac{\lambda }{a}$$ is the distance between adjacent minima.

As $a$ becomes large, we see that the minima will merge together. This is consistent with what we said at the beginning, that if $$a>>\lambda$$ 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 $L$ then we have only considered the case that $L>>\lambda$ and so diffraction occurs only in the other dimension.