Diffraction and Gates

Diffraciton and Gates

Diffraction

Single-slit Diffraction

The intensity in single-slit diffraction is

\[ \begin{gathered} \Delta \varphi = \frac{2\pi}{\lambda} \frac{a}{N} \sin\theta \\\\ \delta = N \Delta \varphi = \frac{2\pi a\sin\theta}{\lambda} \\\\ \alpha = \frac{\delta}{2} = \frac{\pi a\sin\theta}{\lambda} \\\\ I_{\theta} = E_{\theta}^2 = E_m^2 \left( \frac{\sin \alpha}{\alpha} \right)^2 = I_m \left( \frac{\sin \alpha}{\alpha} \right)^2 \end{gathered} \]

When \(\alpha = m\pi\), that is \(a\sin\theta = m\lambda\), \(I_{\theta} = 0\), which is the minimum. The half-angle width for the bright fringe at the center takes \(m = 1\), that is, \(\Delta \theta \approx \sin\theta = \lambda/a\), \(\Delta y_m \approx f\cdot \Delta \theta = f\lambda/a\). So \(a\) is smaller, \(\Delta \theta\) and \(\Delta y_m\) is bigger.

When \(\theta = 0\), the intensity is the maximum. As for others, we have:

\[ \begin{gathered} \frac{d}{d\alpha} \left( \frac{\sin \alpha}{\alpha} \right) = 0 \quad \Rightarrow \quad \alpha = \tan \alpha \\\\ \Rightarrow \alpha = \pm 1.43\pi, \pm 2.46\pi, \pm 3.47\pi, \cdots \\\\ \frac{I_1}{I_m} = 0.045, \quad \frac{I_2}{I_m} = 0.016, \quad \frac{I_3}{I_m} = 0.0083 \end{gathered} \]

All in all,

\[ \delta = a\sin\theta = \begin{cases} 0 & \text{Bright fringe at the center} \\ 2m \cdot \dfrac{\lambda}{2} & m = \pm 1, \pm 2, \pm 3, \cdots \text{ minima} \\ (2m + 1) \cdot \dfrac{\lambda}{2} & m = \pm 1, \pm 2, \pm 3, \cdots \text{ maxima} \end{cases} \]

Fraunhofer Diffraction at Circular Aperture and Angular Resolution

The intensity iof Fraunhofer diffraction is \(\(I(\theta) = I_0 \left( \frac{2J_1(x)}{x} \right)^2, \quad x = \frac{2\pi a\sin\theta}{\lambda}\)\)

where \(J_1(x)\) is the first order Bessel function, \(a\) is the radius of circular aperture.

The half-angle width is approximately \(\Delta \theta = 0.61 \dfrac{\lambda}{a} = 1.22 \dfrac{\lambda}{D}\).

Rayleigh’s criterion: Two objects are just resolved when the maximum of one is at the minimum of the other. \(\theta_R = \theta_{\min} = 1.22 \dfrac{\lambda}{D}\), the resolution ability is \(1/\theta_R\).

Gratings

The intensity of diffraction for \(N\) slits is \(\(E_1 = E_m \left( \frac{\sin \alpha}{\alpha} \right)^2 e^{i0}, \cdots, E_N = E_m \left( \frac{\sin \alpha}{\alpha} \right)^2 e^{i(N-1)\delta}\)\)

where \(\delta = 2\pi/\lambda \cdot d\sin\theta = 2\pi d\sin\theta/\lambda\) is the phase angle between adjacent waves.

\[ \begin{gathered} \beta = \frac{\delta}{2} = \frac{\pi d\sin\theta}{\lambda} \\\\ E_{\theta} = 2\cdot \frac{E_1}{2\sin\beta}\cdot \sin N\beta = E_1 \frac{\sin N\beta}{\sin \beta} \\\\ \Rightarrow I_{\theta} = I_m \left( \frac{\sin \alpha}{\alpha} \right)^2 \left( \frac{\sin N\beta}{\sin \beta} \right)^2 \end{gathered} \]

where \(d = a + b\) is the distance between slits.

If \(\sin\beta = 0\), we have \(\lim\limits_{\sin\beta\rightarrow 0} \dfrac{\sin N\beta}{\sin\beta} = N\), \(I_{\theta} = N^2 I_m\).

If \(\sin N\beta = 0\), we have \(\beta = (m + \dfrac{n}{N}) \pi\), \(\sin\theta = \dfrac{\lambda}{d} (m + \dfrac{n}{N})\). So between two principal maxima, there are \(N-1\) minima and \(N-2\) maxima.

As for the half-angle width of a main maximum, we have

\[ \begin{gathered} a\sin\theta = \lambda \\\\ \Rightarrow Nd\cos\theta \Delta \theta = \lambda \\\\ \Rightarrow \Delta \theta = \frac{\lambda}{Nd\cos\theta} \approx \frac{\lambda}{Nd} \end{gathered} \]

Dispersion and resolving power

Define the dispersion power as the angular seperation per unit wavelength internal:

\[ \begin{gathered} d\cos\theta \cdot \Delta \theta = m \Delta \lambda \\\\ \Rightarrow D = \frac{\Delta \theta}{\Delta \lambda} = \frac{m}{d\cos\theta} \end{gathered} \]

Define the resolving power as the ability to spatially separate two wavelengths. By the Rayleigh’s criterion, the half-angular width \(\Delta \theta_w\) should be resolvable. Therefore, the smallest difference in wavelengths that can be distinguished at a wavelength of \(\lambda\) is \(\(\Delta \lambda = \frac{\Delta \theta_w}{D_{\theta}} = \frac{\lambda}{Nd\cos\theta} \frac{d\cos\theta}{m} = \frac{\lambda}{Nm}\)\)

So the resolving power at \(m\)th order is \(R = \dfrac{\lambda}{\Delta \lambda} = Nm\).