Section Physical Optics (Diffraction and Diffraction Grating)
For a single-slit of width $a$, destructive interference occurs at the angle give by:
Compare with the double-slit inteference side by side:
| Single-slit | Double-slit |
|---|---|
| $a \sin\theta_m = m \lambda$ | $d \sin\theta_m = m \lambda$ |
| $a$: width of the slit | $d$: separation between two slits |
| minima for integer $m\neq 0$ | maxima for integer $m$ |
| $m=0$ gives the central maximum. | $m=0$ gives the central maximum. |
| central maximum twice as wide as other maxima | central maximum is approximately the same width as other maxima |
| central maximum much brighter than other maxima | central maximum is roughly the same brightness as other maxima (for very narrow slits) |
A lab manual based on this simulation is available here.
Drag on the barriers to change slit width. Drag the screen and the circle to change their positions.
The phasor diagram can also be moved by dragging on the lime-colored dot.
Red/blue are used to represent long/short wavelength for illustrative purpose only. The numerical
values of the wavelength in this simulation do not in fact match with the actual wavelength of
visible light.
Applying the small angle approximation, we have: $$ \begin{eqnarray} y_m &=& L \tan \theta_m \approx L \sin \theta_m \tag{*} \end{eqnarray} $$
But we know the $m$-order minimum obeys: $$ \begin{eqnarray} a\sin\theta_m &=& m\lambda \\ \Rightarrow \sin \theta_m &=& m \frac{\lambda}{a} \end{eqnarray} $$ Put this into (*) gives: $$ \begin{eqnarray} y_m &\approx& L \sin \theta_m \\ &=& L (m \frac{\lambda}{a}) \\ \Rightarrow y_m &\approx& m \frac{\lambda}{a} L \end{eqnarray} $$ Remember $m\neq 0$ for the minima.
For single-slit, because the central maximum is so much brighter than all other maxima, we usually only care about the central maximum. The dark fringes on both sides of the central maximum is $m=\pm 1$: $$ \begin{eqnarray} \theta_1 &\approx& \sin \theta_1 = \frac{\lambda}{a} \\ y_1 &\approx& \frac{\lambda}{a} L \end{eqnarray} $$
The width $w$ of the central maximum is twice that of $y_1$: $$ w = 2 y_1 = 2\frac{\lambda}{a} L $$
$y_m$ without small angle approximationFor waves of long wavelengths, such as sound, radio waves, microwaves, one should not use the small angle approximation (because of large angle $\theta$). In this case, the minima are at ($m\neq 0$): $$ \begin{eqnarray} a\sin\theta_m &=& m\lambda \\ \Rightarrow \theta_m &=& \sin^{-1} (m \frac{\lambda}{a}) \end{eqnarray} $$ Putting into $y_m = L \tan \theta_m$ gives: $$ y_m = L \tan (\sin^{-1} (m \frac{\lambda}{a})) $$ Most of the problems in this course are for visible light, so it is usually safe to make the small angle approximation.
Since the first minimum is at angle $\theta_1 \approx \frac{\lambda}{a}$, Rayleigh's criterion requires the angular separation $\theta$ to obey:
For a circular aperture (as opposed to a slit), the minimum angle is modified by a factor to $\theta_1 \approx 1.22 \frac{\lambda}{a}$, hence:
A lab manual based on this simulation is available here.
Drag on the barriers to change slit width. Drag the screen and the circles (light sources) to change their positions.
Locations of the maxima of a diffraction grating is the same as the double-slit, given by:
⚠ $d$ for diffration gratings could be made very small, giving relatively large value of $\theta$ ($\theta \sim 1rad$). Therefore it is better not to use the small angle approximations.