Question

Light with wavelength $\lambda$ falls normally on a long rectangular slit of width $b$. Find the angular distribution of the intensity of light in case of Fraunhofer diffraction, as well as the angular position of the minima.

Answer 1

If a plane wave is incident normally from the left on a slit of width b and the diffracted wave is observed at a large distance, the resulting pattern is called Fraunhofer diffraction. The condition for this is $b^2<<l\lambda$ where l is the distance between the slit and the screen. In practice light may be focused on the screen with the help of a lens (or a telescope).

Consider an element of the slit which is an infinite strip of width dx. We use the formula of problem 5.103 with the following modifications.

The factor $\frac{1}{r}$ characteristic of spherical waves will be omitted. The factor $K\left(\phi \right)$ will also be dropped if we confine overselves to not too large $\phi$. In the direction defined by the angle $\phi$ the extra path difference of the wave emitted from the element at x relative to the wave emitted from the centre is $\Delta =-x\sin \phi$

Thus the amplitude of the wave is given by $\alpha {\int }_{-b/2}^{+b/2}{e}^{ik\sin \phi }dx=(e^{i\frac{1}{2}kb\sin \phi }-e^{-1\frac{1}{2}kb\sin \phi })/ik\sin \phi$

$=\frac{\sin \left(\frac{\pi b}{\lambda }\sin \phi \right)}{\frac{\pi b}{\lambda }\sin \phi }$

This $I=I_0\frac{{\sin }^2\alpha }{{\alpha }^2}$

where $\alpha =\frac{\pi b}{\lambda }\sin \phi$ and $I_0$ is a constant

Minima are observed for $\sin \alpha =0$ but $a\approx 0$

Thus we find minima at angles given by $b\sin \phi =k\lambda ,k=\pm 1,\pm 2,\pm 3,....$