Question

Draw a graph to show the relative intensity distribution for a single slit diffraction pattern. Obtain the expression for the width of central maxima. If slit width is doubled then what is the effect on width of central maxima?

Answer 1

Let the width of the slit be through which diffraction pattern is obtained be $b$ and $\lambda$ be the wavelength of light used.

The first minima of diffraction pattern is formed at point P on one side and at A of other side.

Path difference between the light rays reaching at point P $\Delta x=S_{2}P-S_{1}P$

$\therefore$ $\Delta x=\frac{b}{2}\sin \theta$

For first minima, path difference must be equal to $\frac{\lambda }{2}$

$\therefore$ $\frac{b}{2}\sin \theta =\frac{\lambda }{2}$

OR $b\sin \theta =\lambda$

As $\theta$ is very small, thus $\sin \theta \approx \tan \theta =\frac{y}{D}$

OR $b\times \frac{y}{D}=\lambda$ $⟹y=\frac{\lambda D}{b}$

Width of central maxima $PA=2y=\frac{2\lambda D}{b}$

$⟹$ $PA\propto \frac{1}{b}$

Hence the width of central maxima formed becomes half of its initial value as the slit width is doubled.