Physics 2 Lecture 26

Diffraction

• Diffraction pattern is derived from the interference of all wavelets using the amplitude and phase difference of each.
• Interference is superposition of a “few” waves, diffraction is used when there are many waves.
• Diffraction problems have two cases:
  • Fraunhofer (Far-Field)
  • Fresnel (Near-Field)
• Relative length scale set by $\lambda$ , size of aperture, and distance to observer.
• For a circular aperture phase difference from center to edge is given by

$$\Delta \varphi =\frac{2\pi }{\lambda }\left[\sqrt{d^2+\frac{a^2}{4}}-d\right]=\frac{2\pi }{\lambda }d\left[\sqrt{1-\frac{a^2}{4dsr}}-1\right]\approx \frac{2\pi }{\lambda }\left(\frac{a^2}{8d}\right)$$

• If $d\gg \frac{a^2}{\lambda }$ then it is reasonable to treat rays as parallel and $\Delta \varphi$ to be small.
• If $d<\frac{a^2}{\lambda }$ then the phase differences become large and rays are not parallel.

Diffraction as N-source Interference

Assuming:

• No initial phase differences
• Sources are small compared to wavelength
• Observation point P is far away The amplitudes of waves arriving at P are equal

$$E_0(r_1)=E_0(r_n)$$

The phase difference between adjacent sources is given by:

$$\frac{2\pi }{\lambda }d\sin \theta$$

And maxima occurs when:

$$d\sin {\theta }_m=m\lambda$$

Diffraction gratings are modeled by this:

$$m\Delta \lambda =d\cos \theta \Delta \theta ⟹\Delta \theta =\frac{m\Delta \lambda }{d\cos \theta }$$

Two different wavelengths show principle maxima @ different values of ${\theta }_m$. Thus, diffraction gratings are used to separate wavelengths.

The width of a “strong peak” is approx $\delta \theta =\frac{\lambda }{Nd\cos \theta }$ where N = number of slits.

Single Slit Diffraction

We can model a single slit diffraction as $n$ wavelets emitting $n$ parallel rays at an angle $\theta$. Dividing the sources into two zones of width $\frac{a}{2}$, then the angle of the first intensity minimum occurs when destructive interference between all of zone 1 and 2. First Destructive interference of a slit with width $a$ occurs when

$$\frac{a}{2}\sin {\theta }_1=\frac{\lambda }{2}$$

This can be generalized to the $m$‘th slit by dividing the slit width with larger numbers. Rearranging this idea, we get the formula below:

$$a\sin {\theta }_m=m^{\prime }\lambda$$

where

$$m^{\prime }=\pm 1,\pm 2,\pm 3\dots$$

Intensity of single slit diffraction

Intensity is obtained by summing all the wavelets from the $n$ rays to get the amplitude. This can be modeled with phasors.