It turns out that figuring out intensity in the diffraction pattern is not so easy as finding the directions of the minima. We must apply the superposition principle to the waves, and then, find the time-average of the square of the sum.
We will use Figure 48.9 to perform out calculations. I have placed a \(z\) axis at the slit to refer to parts of the wavefront there. Let \(y\) axis be pointed into the page and \(x\) axis pointed to the right. Note that this figure shows only a cross-section of the wavefront at the slit since the wavefront is parallel to the \(yz\) plane.
Figure48.9.Geometry for the calculation of Fraunhofer diffraction from plane wave at a slit. We assume electric field to be mostly along \(z\) axis. Here \(\hat u_z\) is unit vector towards positive \(z\) axis. \(dz\) shows a small part of the wavefront at coordinate \(z\text{.}\)
We will also assume that electric field is mostly the \(z\) component. Therefore, work with just the \(E_z\) wave and call \(E_z\) by \(E\text{.}\) Let \(A\) denote the voltage per unit length at the wavefront. Then, we can write the electric field wave at point P on the screen which started at \(dz\) of the wavefront as a spherical wave as follows.
\begin{equation}
dE = \frac{A dz}{r}\cos(kr - \omega t),\tag{48.3}
\end{equation}
\(Adz/r\) is the amplitude of this wave when it arrives at P, \(k\) is the wavenumber, which is related to wavelength \(\lambda\) by
\begin{equation*}
k = \frac{2\pi}{\lambda},
\end{equation*}
\(\omega\) is the angular frequency, which is related to the frequency \(f\) by
\begin{equation*}
\omega = 2\pi f.
\end{equation*}
Simplifying \(r\text{:}\)
From triangle, whose vertices are O, P, and \(dz\text{,}\) we get the following for \(r^2\text{.}\)
Here since \(|z|\lt \lt R\text{,}\) we can expand the quantity in parenthesis and keep upto two terms so that we have dependence on \(z\) still there.
\begin{equation*}
r = \frac{R}{\cos\theta} - z \sin\theta.
\end{equation*}
We make one more simplification here. Since \(\theta\) will be small, we can replace \(\cos\theta\) by \(1\text{.}\) To be consistent, we could replace \(\sin\theta\) by \(\theta\) in radians, but, it is customary to leave \(\sin\theta\) alone here. Finally, we have the following simplified expression for \(r\text{.}\)
\begin{equation}
r = R - z \sin\theta.\tag{48.4}
\end{equation}
Electric Field at P:
In Eq. (48.3), distance \(r\) appears in the amplitude \(\frac{Adz}{r}\) and in phase \(kr - \omega t\text{.}\) Since \(|z\sin \theta| \lt\lt R\) we may be tempted to drop \(z\sin \theta\) altogether. Since \(kr\) in the phase has a large \(k\text{,}\) small changes in \(kr\) have large impact on sine of \(kr\) . Therefore, in the phase we must keep \(z\sin \theta\text{.}\) Dropping \(z\sin \theta\) in the the amplitude \(\frac{Adz}{r}\) does not have much impact. Thus, Eq. (48.3) simplifies to
\begin{equation*}
dE \approx \frac{(Adz)}{R}\sin\left[ k(R-z\sin\theta) - \omega t\right]
\end{equation*}
Now, we just integrate this over the wavefront, from \(z=-b/2\) to z=+b/2, to get the electric field at P which is in the direction \(\theta\) from the slit.
Intensity at in the direction \(\theta\) is proportional to time average of this fluctuating field. Since time averaging of \(\sin^2(\omega t)\) and \(\cos^2(\omega t)\) give \(\frac{1}{2}\text{,}\) after calculations, we end up with the following for intensity.
\begin{equation}
I(\theta)=\epsilon_0 c \left( \frac{bA}{R} \right)^2 \left[ \frac{\sin\beta}{\beta} \right]^2,\tag{48.6}
\end{equation}
where symbol \(\beta\) stands for
\begin{equation}
\beta = \frac{\pi b \sin\theta}{\lambda}.\tag{48.7}
\end{equation}
Let us denote the intensity in the \(\theta_0\) direction by \(I_0\text{.}\) Since \(sin x/x = 1 \) when \(x=0\text{,}\) from Eq. (48.6) we get
\begin{equation}
I_0 = \epsilon_0 c \left( \frac{bA}{R} \right)^2.\tag{48.8}
\end{equation}
This helps us write intensity in direction \(\theta\) in terms of \(I_0\text{.}\) We rewrite Eq. (48.6) as
We now use the formula for intensity (48.9) to find the values of \(\beta\) in which intensity will have maxima and minima. After we get these \(\beta\) values, we use Eq. (48.7) to obtain angles \(\theta\) which give us the directions from the center of the slit.
Diffraction Minima
First we notice that when \beta = 0, \(sin\beta /\beta\) is not zero, but it is equal to 1. For \(\beta = m^\prime \pi\) for non-zero integer \(m\text{,}\)\(\sin\beta /\beta\) is zero. Therefore, we have all the betas for which we will have zero intensity, which are the minima here. Let us give subscripts to these betas.
The intensity maxima do not occur at points halfway between the minima. Instead we must find \(\beta\) values at the maxima of intensity \(I\) by treating \(I(\beta)\) as a function of \(\beta\text{.}\) By setting the derivative \(dI/d\beta\) to zero, we will get both maxima and minima, and since we already know the minima, we will identify the maxima by removing the minima from the list. \(dB/d\beta=0\) gives
The first condition includes one maximum at \(\beta = 0\text{,}\) and the rest of the solutions are minima as discussed above.
The second condition corresponds to only maxima. You can verify that this is the case by using second derivative test. Now, you should notice that, unlike solving \(\sin\beta = 0\text{,}\) solving \(\beta\cos\beta -\sin\beta = 0\) is not an easy task. This equation is a transcendental equation, which does not have a nice close-form solutions, but they can be solved graphically as we will show below.
Before, we solve Eq. (48.14), by dividing by \(\cos\theta\) we convert this equation to the following since it is easier to see when \(\tan\beta\) and \(\beta\) cross than when \(\beta\cos\beta -\sin\beta = 0\) and \(\sin\beta\) cross.
We now plot the two sides of the equation as two functions, \(\tan \beta\) and \(\beta\text{,}\) versus \(\beta\) as shown in Figure 48.10. We have a graphical solution wherever the two graphs intersect since we are looking for them to be equal. Of course you can write a program to get the nearest such point to your desired precision. You can also read off the values from the graph for a rough answer: for instance, 0, \(\pm\frac{3}{2}\pi\text{,}\)\(\pm\frac{5}{2}\pi\text{,}\) etc.
Figure48.10.The graphical solution of \(\tan\beta = \beta.\text{.}\)
A more precise and expanded graph gives solutions to be
The corresponding direction angles from the center of the slit will be
\begin{align}
\theta_m \amp = \sin^{-1}\left(\frac{\beta_m}{\pi}\: \frac{\lambda}{b}\right)\tag{48.15}\\
\amp \approx
\begin{cases}
0 \amp \text{ for } m = 0,\\
\pm \frac{2 m + 1}{2} \frac{\lambda}{b} \amp \text{ for } m = 1, 2, 3, \cdots.
\end{cases}\tag{48.16}
\end{align}
Intensity Plot
We can plot \(I/I_0\) versus \(\beta\) or versus \(\theta\text{.}\) When you plot \(I/I_0\) versus \(\beta\text{,}\) the minima will occur at non-zero integer multiples of \(\pi\text{,}\) which correspond to \(\theta\) between \(-\pi/2\) and \(\pi/2\text{.}\)
Figure 48.11 shows three plots of \(I/I_0\) correponding to three cases of slit width compared to wavelength: \(b=\lambda/2\text{,}\)\(b=\lambda\text{,}\) and \(b=2\lambda\text{.}\) The \(b=\lambda/2\) is too narrow a slit to have diffraction minima and shows what to expect if you have very narrow opening. The \(b=2\lambda\) shows five diffraction maxima, and \(b=\lambda\) is the borderline case which just shows noticeably pointed spread in intensity and no minima.
Figure48.11.A plot of diffraction pattern for various slit widths. Only with slit width greater than the wavelength that we would observe diffraction maxima and minima.
It is worth noting that the intensity of the central peak is about 20 times the intensity of the first side peaks. Evaluating \(I/I_0\) at the peaks, we find
If you plotted \(I/I_0\) versus \(\beta\text{,}\) the central peak will go from \(\beta=-\pi\) to \(\beta=\pi\) at the base. That is the central peak covers a range of \(\Delta\beta=2\pi\text{.}\) The other peaks cover half as wide a range. For instance, the first side peak goes from \(\beta=\pi\) to \(\beta=2\pi\text{,}\) which gives it a range of \(\Delta\beta=\pi\) only.
These ranges in \(\beta\) values can be converted to corresponding ranges in \(\theta\) values. For the central peak, we will get the \(\theta\) ranage to be
\begin{align*}
\amp \text{Central: }\ \theta \text{ from } -\sin^{-1}{\lambda/b} \text{ to } \sin^{-1}{\lambda/b},\\
\amp \text{Next peak on right: }\ \theta \text{ from } \sin^{-1}{\lambda/b} \text{ to } \sin^{-1}{2\lambda/b},\text{ etc.}
\end{align*}
For small angles, these become
\begin{align*}
\amp \text{Central: }\ \theta \text{ from } - \lambda/b \text{ to } \lambda/b \Longrightarrow \Delta\theta=2\lambda/b,\\
\amp \text{Next peak on right: }\ \theta \text{ from } \lambda/b \text{ to } 2\lambda/b \Longrightarrow \Delta\theta=\lambda/b,\text{ etc.}
\end{align*}
Thus, in small angle case, we do see that the base of central peak is twice as wide as for other peaks. Although, width of a peak is defined at the half height, not at the base, we still get this two-to-one ratio between the width of the central maximum versus the widths of other peaks as far as width in \(\beta\) is concerned.
Example48.12.Widths and Positions of Diffraction Through a Single Wide Slit.
Monochromatic light of wavelength 530 nm passes through a horizontal single slit of width 1.5 \(\mu\)m in an opaque plate. A screen of dimensions 2 m \(\times\) 2 m is 1.2 m away from the slit. (a) Which way is the diffraction pattern spread out on the screen? (b) What are the directions of the minima? (c) What are the directions of the maxima? (d) How wide is the central bright fringe on the screen? (e) How wide is the next bright fringe on the screen?
Since the slit is horizontal, the diffraction pattern will be vertical.
Solution2.(b)
The diffraction minima from a single slit occur when \(\beta = m \pi\) for \(m = \pm1,\: \pm 2,\: \cdots\text{.}\) In terms of the slit width \(b\) and the direction \(\theta\) vertically from the slit this condition becomes
\begin{equation*}
\dfrac{\pi b \sin\:\theta_m}{\lambda} = m \pi,\ \ \Longrightarrow\ \ b\:\sin\:\theta_m = m\:\lambda,\ \ m = \pm1,\: \pm 2,\: \cdots.
\end{equation*}
Using this formula we obtain the following values for \(\theta_1\text{,}\)\(\theta_2\) and \(\theta_3\text{.}\)
That means \(\theta_3\) does not exist. Therefore, the positions of the first and second order spots on the screen will be given by the \(y\)-coordinates,
Example48.13.Width of a Single Slit from Width of Central Maximum on Screen.
A monochromatic light of wavelength \(632.8\, \text{nm}\) incident on a single slit has central bright spot of width \(5\, \text{cm}\) at a screen \(1.6\, \text{m}\) away. Find the width of the slit.
Answer.
\(40.5\ \mu\)m.
Solution.
From the given data we can find the direction to the \(m=1\) minimum.
\begin{equation*}
b = \dfrac{632.8\:\text{nm}}{\sin\:0.895^{\circ}} = 40.5\:\mu\text{m}.
\end{equation*}
Example48.14.Separation of Two Wavelengths in a Single-slit Diffraction.
A source of light having two wavelength \(550\, \text{nm}\) and \(600\, \text{nm}\) of equal intensity is incident on a slit of width \(1.8\, \mu\)m. Find the separation of the \(m = 1\) bright spots of the two wavelengths on a screen \(30\, \text{cm}\) away.
Answer.
\(2.1\, \text{cm}\text{.}\)
Solution.
From the condition for the diffraction maxima we have \(b\:\sin\:\theta = 1.4303\:\lambda\) for \(m=1\) maximum. Therefore, we would get following angles for the two wavelengths.
Therefore the peaks of the spots are separated by \(\Delta y = 2.1\,\text{cm}\text{.}\)
Example48.15.Intensity of Light in a Particular Direction behind a Single Slit.
A single slit of width \(3.0\, \mu\text{m}\) is illuminated by a sodium yellow light of wavelength \(589\, \text{nm}\text{.}\) Find the intensity at a \(15^{\circ}\) angle to the axis in terms of the intensity of the central maximum which can be taken to be \(1\) unit.
Answer.
\(0.04\)
Solution.
The intensity in the diffraction pattern of a single slit is given by
