Intensity of central maxima of crystal diffraction

[phantomvommand]

Homework Statement

Please see attached photo.

Relevant Equations

Intensity proportional to Amplitude of E-field squared.

When the crystal above is illuminated with light of intensity $I_0$, what is the intensity at the central maximum? (The picture shown above is a 4 x 4 unit cell of the crystal)
The answer is $(\frac {5} {16} )^2 I_0$. Why?

Apparently, Electric field is proportional to the transparent area, and will be maximum if the whole unit cell is transparent. I feel that it should be intensity that's proportional to the transparent area instead?
I cannot see how amplitude of electric field is related to transparent squares.

Thanks for all the help.

 

[Charles Link]

From diffraction theory, the intensity of the primary maxima for $N$ sources (in phase), (or for a single slit diffraction pattern made from $N$ sources), is proportional to $N^2$, because the resultant electric field amplitude $E$ is proportional to $N$. The reason energy is conserved is that for $N$ sources, the angular width of the diffraction pattern (central maximum) $\Delta \theta \approx \frac{\lambda}{b}$, where $b$ is the width of the $N$ sources, (e.g. arranged in a square), so that the solid angle of the maximum, which is proportional to $(\Delta \theta )^2$, is inversely proportional to the total area of the sources. Thereby the solid angle of the diffraction maximum is inversely proportional to $N$. The power is a product of the intensity and the solid angle. It turns out the power in the entire maximum for $N$ sources will be proportional to $N$. The peak goes as $N^2$, but the solid angle spread goes as $\frac{1}{N}$.

Note: With circular symmetry, (with a circular source of diameter $d$), the $\Delta \theta$ of the diffraction pattern, which is an Airy disc pattern, is once again proportional to $\frac{\lambda}{d}$, so that the solid angle of the diffraction maximum is again inversely proportional to the area of the source.

 

[hutchphd]

And the "book" answer is that the forward amplitude is proportional to the corresponding (0,0) Fourier component which is just the normalized integral. But I like the @Charles Link explanation also..

.

 

[Fred Wright]

I analyze this problem using Fraunhofer diffraction theory, to whit, the complex amplitude of the diffracted wave is,
$$
U(z,y,z)\propto \int \int_{Aperture}A(x^{'},y^{'})e^{-i\frac{2\pi}{\lambda}(ux^{'}+vy^{'})}dx^{'}dy^{'}
$$
where $A(x^{'},y^{'})$ is the complex amplitude over the surface of the aperture. I wave my hand and declare the constant of proportionality is $\sqrt{I_0}$, and refer to the diagram below for the limits of integration in the five transparent regions of the unit cell.

$$
U_1= \sqrt{I_0}\int_{-\frac{1}{2}}^{-\frac{1}{4}}e^{-i\frac{2\pi}{\lambda}ux^{'}}dx^{'}\int_{-\frac{1}{2}}^{-\frac{1}{4}}e^{-i\frac{2\pi}{\lambda}vy^{'}}dy^{'}
$$
$$
=\sqrt{I_0}\frac{e^{-i\frac{2\pi}{\lambda}ux^{'}}}{-i\frac{2\pi}{\lambda}u}|_{-\frac{1}{2}}^{-\frac{1}{4}}\frac{e^{-i\frac{2\pi}{\lambda}vy^{'}}}{-i\frac{2\pi}{\lambda}v}|_{-\frac{1}{2}}^{-\frac{1}{4}}
$$
$$
=\sqrt{I_0}\frac{e^{i\frac{\pi}{4\lambda}(3u+3v)}}{16}sinc(\frac{\pi u}{4\lambda})sinc(\frac{\pi v}{4\lambda})
$$
in likewise manner we compute,
$$
U_2=\sqrt{I_0}\frac{e^{i\frac{\pi}{4\lambda}(u+v)}}{16}sinc(\frac{\pi u}{4\lambda})sinc(\frac{\pi v}{4\lambda})
$$
$$
U_3=\sqrt{I_0}\frac{e^{i\frac{\pi}{4\lambda}(u-v)}}{16}sinc(\frac{\pi u}{4\lambda})sinc(\frac{\pi v}{4\lambda})
$$
$$
U_4=\sqrt{I_0}\frac{e^{i\frac{\pi}{4\lambda}(-u+v)}}{16}sinc(\frac{\pi u}{4\lambda})sinc(\frac{\pi v}{4\lambda})
$$
$$
U_5=\sqrt{I_0}\frac{e^{i\frac{\pi}{4\lambda}(-3u-v)}}{16}sinc(\frac{\pi u}{4\lambda})sinc(\frac{\pi v}{4\lambda})
$$
We sum the individual contributions and find the intensity by taking the dot product of $U$ with its complex conjugate.
$$
U=U_1+U_2+U_3+U_4+U_5
$$
$$
I=U\cdot U^{*}
$$
$$
=\frac{I_0}{(16)^2}sinc(\frac{\pi u}{4\lambda})^2sinc(\frac{\pi v}{4\lambda})^2[5+4\cos(\frac{\pi}{2}(u+v))+2\cos(\frac{\pi}{2}(u+2v))+2\cos(\frac{\pi}{2}(2u+v))+2\cos(\frac{\pi}{2}(3u+2v))+2\cos(\frac{\pi}{2}(u))+2\cos(\frac{\pi}{2}(v))+2\cos(\frac{\pi}{2}(u-v))2\cos(\frac{\pi}{2}(4u))+2\cos(\frac{\pi}{2}(4u+2v))]

$$
We evaluate this at the center of the cell to find $I=I_0\frac{(5)^2}{(16)^2}$.