9.1 Double slit diffraction

Two slit diffraction

Now we consider the case of two slit diffraction.

Now we we will define $R=R_1$ and use $R_2=R-d\sin \theta$ $$E=\frac{{\epsilon }_{L}a}{R}sinc\beta e^{i(kR-\omega t)}+\frac{{\epsilon }_{L}a}{R-d\sin \theta }sinc\beta e^{i(kR-kd\sin \theta -\omega t)}$$ Now we can ignore the $d\sin \theta$ in the denominator, as it will not have a significant effect on that. However in the exponent, we can not ignore it, since it could significantly affect thephase of the harmonic function. Lets define $\alpha =(kd\sin \theta )/2$ so now we can write: $$E=\frac{{\epsilon }_{L}a}{R}sinc\beta e^{i(kR-\omega t)}+\frac{{\epsilon }_{L}a}{R}sinc\beta e^{i(kR-2\alpha -\omega t)}$$ and start rearranging: $$$$ $$$$ $$E=\frac{{\epsilon }_{L}a}{R}sinc\beta \left[2\cos \alpha \right]e^{i(kR-\alpha -\omega t)}$$ This is very similar to the case of single slit diffraction except that you now get a factor $2\cos \alpha$ included and a phase shift in the harmonic function.

So we can see immediately the intensity is $$I=4I_0{\cos }^2\alpha sin{c}^2\beta$$ recall $$\alpha =(kd\sin \theta )/2$$ and $$\beta =(ka\sin \theta )/2$$ If $d$ goes to 0 then expression just becomes the expression for single slit diffraction. If a goes to 0 then the expression just becomes that for Youngsdouble slit. The double slit diffraction is just the product of these two results. (Hey cool!)