Question 7

BONUS QUESTION: 5 Points

Do Not Attempt This Question Until After Attempting All Other Questions.

A plane wave of wavelength $\lambda$is normally incident upon a rectangular aperture in an otherwise opaque mask. The rectangular aperture is of length a and width b. Find the intensity beyond the aperture in the Fraunhofer limit.

For this problem, it is easiest to use complex variables to denote the phasor diagram. We divide the region in the mask into small surface elements, and then add up each contribution to the electric vector at each point on the screen.

If the element on the screen is dS, then the contribution to the electric vector in the frounhofer limit is

$$d{\bf E}~=~{{\bf E_m} \over r}\ e^{i(\omega t-{\bf k}\cdot {\bf r})} dS$$

where ${\bf E_m}$ is the electric field per unit area at the mask, r is the distance to the point on the screen, in the direction of ${\bf k}$. We have that

$$r~=~\sqrt{ X^2+(Y-y)^2 +(Z-z)^2 }~=~R- {2 (Yy+Zz)\over R}+...$$

where $R~=~\sqrt{X^2+Y^2+Z^2}$ is the distance from the centre point of the rectangular aperture to the point on the screen we are interested in. The ellipses denote terms higher order in the size of the aperture divided by the diatance to the screen, corrections to the frounhofer limit.

$${\bf E}~=~\int d{\bf E}~=~ \int_{\rm Aperture} \ dS\ {{\bf E_m} \over r}\ e^{i(\omega t-{\bf k}\cdot {\bf r})}$$

$${\bf E}~=~\int^{b\over 2}_{-{b\over 2}} \ dy\ \int^{a\over 2}_{-{......z{{\bf E_m} \over R}\ e^{i(\omega t- k\cdot {\bf R})}e^{i k {(Yy+Zz)\over R}}$$

which factors nicely into

$${\bf E}~=~ {{\bf E_m} \over R}\ e^{i(\omega t- k\cdot {\bf R})}\...... dy\ e^{i kY{y\over R}}\int^{a\over 2}_{-{a\over 2}} \ dz e^{i k Z {z\over R}}$$

Now we use the simple integration

$$\int^{b\over 2}_{-{b\over 2}} \ dy\ e^{i kY{y\over R}}~=~b \left({\sin\beta\over\beta}\right)$$

where

$$\beta~=~{k b Y \over 2R}$$

to find

$${\bf E}~=~ {{\bf E_m} a b \over R}\ e^{i(\omega t- k\cdot {\bf R})}\left({\sin\beta\over\beta}\right) \left({\sin\alpha\over\alpha}\right)$$

where

$$\alpha~=~{k a Z \over 2R}$$

We therefore find that

$$I(Y,Z)~=~I(0,0)~=~ \left( {\sin\beta\over\beta}\right)^2\left( {\sin\alpha\over\alpha}\right)^2$$

which reproduces our usual result when we take $b\rightarrow\infty$, for diffraction from a single slit, by identifying

$$\alpha~=~{k a Z \over 2R}~=~{\pi a\sin\theta\over\lambda}$$