- #1
- 1,097
- 1,384
- Homework Statement
- Determine the factor by which the on-axis intensity from a square aperture of side length ##a## is reduced, compared to an unobstructed pattern.
- Relevant Equations
- Fresnel integral.
I'd appreciate if someone could check whether my work is correct. The ##x##-##y## symmetry of the aperture separates the Fresnel integral:\begin{align*}
a_p \propto \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{ikx^2}{2R} \right) dx \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{iky^2}{2R} \right) dy \equiv \left[ \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{ikx^2}{2R} \right) dx \right]^2
\end{align*}I substitute ##\frac{\pi \xi^2}{2} = \frac{kx^2}{2R}##, i.e. ##\xi = x\sqrt{\frac{k}{\pi R}}##. I also denote ##w = \frac{a}{2} \sqrt{\frac{k}{\pi R}}## to simplify the new limit,\begin{align*}
a_p \propto \frac{\pi R}{k} \left[ \int_{-w}^{w} \mathrm{exp}\left(\frac{i\pi \xi^2}{2} \right) d\xi \right]^2 = \frac{4\pi R}{k} \left[ \int_{0}^{w} \mathrm{exp}\left(\frac{i\pi \xi^2}{2} \right) d\xi \right]^2
\end{align*}and separate into sines and cosines,\begin{align*}
a_p \propto \frac{4\pi R}{k} \left[ \int_{0}^{w} \cos{\left(\frac{\pi \xi^2}{2} \right)} d\xi + i\int_{0}^{w} \sin{\left(\frac{\pi \xi^2}{2} \right)} d\xi \right]^2 = \frac{4\pi R}{k} \left[ C\left( w \right) + iS\left( w \right) \right]^2
\end{align*}The unobstructed pattern would instead correspond to ##w \rightarrow \infty##, in which case ##[C(\infty) + iS(\infty)]^2 = [1 + i]^2 = 2i##. The amplitude of the pattern with the square is therefore reduced by a factor\begin{align*}
\alpha = \frac{\left[ C\left( w \right) + iS\left( w \right) \right]^2}{2i}
\end{align*}and the intensity by ##|\alpha|^2##,\begin{align*}
|\alpha|^2 = \frac{|C\left( w \right) + iS\left( w \right) |^4}{4}
\end{align*}Does it look correct?
a_p \propto \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{ikx^2}{2R} \right) dx \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{iky^2}{2R} \right) dy \equiv \left[ \int_{-a/2}^{a/2} \mathrm{exp}\left(\frac{ikx^2}{2R} \right) dx \right]^2
\end{align*}I substitute ##\frac{\pi \xi^2}{2} = \frac{kx^2}{2R}##, i.e. ##\xi = x\sqrt{\frac{k}{\pi R}}##. I also denote ##w = \frac{a}{2} \sqrt{\frac{k}{\pi R}}## to simplify the new limit,\begin{align*}
a_p \propto \frac{\pi R}{k} \left[ \int_{-w}^{w} \mathrm{exp}\left(\frac{i\pi \xi^2}{2} \right) d\xi \right]^2 = \frac{4\pi R}{k} \left[ \int_{0}^{w} \mathrm{exp}\left(\frac{i\pi \xi^2}{2} \right) d\xi \right]^2
\end{align*}and separate into sines and cosines,\begin{align*}
a_p \propto \frac{4\pi R}{k} \left[ \int_{0}^{w} \cos{\left(\frac{\pi \xi^2}{2} \right)} d\xi + i\int_{0}^{w} \sin{\left(\frac{\pi \xi^2}{2} \right)} d\xi \right]^2 = \frac{4\pi R}{k} \left[ C\left( w \right) + iS\left( w \right) \right]^2
\end{align*}The unobstructed pattern would instead correspond to ##w \rightarrow \infty##, in which case ##[C(\infty) + iS(\infty)]^2 = [1 + i]^2 = 2i##. The amplitude of the pattern with the square is therefore reduced by a factor\begin{align*}
\alpha = \frac{\left[ C\left( w \right) + iS\left( w \right) \right]^2}{2i}
\end{align*}and the intensity by ##|\alpha|^2##,\begin{align*}
|\alpha|^2 = \frac{|C\left( w \right) + iS\left( w \right) |^4}{4}
\end{align*}Does it look correct?
