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From the explanation on diffraction at http://web.mit.edu/viz/EM/visualizations/coursenotes/modules/guide14.pdf

$$I = I_0\left [ \frac{\sin(\beta/2)}{\beta/2} \right ]^2=I_0\left [ \frac{\sin(\pi a \sin \theta/\lambda)}{\pi a \sin \theta/\lambda} \right ]^2$$ Where $a$ is the slit width, $\beta = 2\pi a \sin \theta /\lambda$ is the total phase difference between waves from the upper end and lower end of the slit, and $I_0$ is the intensity at $\theta=0$

I did not find on textbooks an answer to my question, which is: how can I calculate $I_0$?

Is it simply equal to the total intensity of the beam incident on slit or is it something different?

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A little bit of background: As stated in your notes, this equation is obtained using the Fraunhofer diffraction formula (which is an approximation to the more general Rayleigh-Sommerfeld formula). According to the Fraunhofer diffraction integral, given the transmittance function of the diffractive screen: $$U(x',y')=\cases{1\quad\quad\text{on the aperture}\\0\quad \quad\quad \text{otherwise}}$$

the diffracted field observed on a plane located at a distance $z$ from the screen would be $$U_O(x,y) = \frac{e^{jkz}e^{j\frac{k}{2z}(x^2+y^2)}}{j\lambda z}\mathcal F\{U_{inc}(x',y')\}|_{(f_x,f_y) }$$ which is just the Fourier transform of the transmittance function, $\mathcal F\{U(x',y')\}$, calculated at the frequencies $(f_x,f_y)=(\dfrac{x}{\lambda z},\dfrac{y}{\lambda z})$, multiplied by some factor. Therefore, within the scope of validity of the Fraunhofer (or far-field) approximation, you just have to Fourier transform the transmittance function in order to find the diffraction pattern.

                                enter image description here

In your problem, we have a single slit with a width of $w_x$ illuminated by a normally incident plane-wave with amplitude A and intensity $ A^2$. Therefore, the transmittance function would be a rectangular function $\mathrm {rect}(\dfrac{x'}{w_x})$. looking up its Fourier transform in some table, we find the diffracted field using the above formula: $$U(x) = A\frac{e^{jkz}e^{j\frac{k}{2z}x^2}}{j\lambda z}\times w_x\mathrm {sinc}(\frac{w_xx}{\lambda z})$$ and for the intensity: $$\boxed{I(x)=|U(x)|^2=A^2\frac{w_x^2}{\lambda ^2 z^2}\mathrm {sinc}^2(\frac{w_xx}{\lambda z})}$$

where $\mathrm {sinc}(x) = \frac{\sin (\pi x)}{\pi x}$. This is essentially the same formula in your question, because in the Fraunhofer approximation you can use $\sin\theta\approx \tan \theta= \dfrac{x}{z}$.

Therefore, $I_0$ in your question would be $A^2\dfrac{w_x^2}{\lambda ^2 z^2}$.

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  • $\begingroup$ So if we make the width of the slit $w_x$ large enough, we get that $I_0$ is actually greater than $A^2$, or in other words, the intensity of the incident plane wave? $\endgroup$ – thermomagnetic condensed boson Jun 24 '17 at 18:02
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    $\begingroup$ @no_choice99 Good point! The answer is obviously no ,we can't do that. This formula is obtained based on the Fraunhofer approximation. One of the assumptions of this approximation is $z\gg\dfrac{k(x'^2+y'^2)_{max}}{2}$ or in this case, $z\gg\dfrac{k(w_x/2)^2}{2}$. Therefore, we will always have $\dfrac{w_x^2}{\lambda^2z^2}<1$. $\endgroup$ – Mo_ Jun 24 '17 at 18:16

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