For a parallel beam of monochromatic light of wavelength $\lambda$, diffraction is produced by a single slit w

Question

For a parallel beam of monochromatic light of wavelength $\lambda$, diffraction is produced by a single slit whose width $a$ is much greater than the wavelength of the light. If $D$ is the distance of the screen from the slit, the width of the central maxima will be:

Accepted Answer

In a single slit diffraction pattern, the angular spread of the central maximum is bounded by the first minima on either side. The condition for the first minimum is $a \sin 	heta = \lambda$. Since $	heta$ is small, $\sin 	heta \approx 	heta = \frac{\lambda}{a}$.
The angular width of the central maximum is $2	heta = \frac{2\lambda}{a}$.
The linear width of the central maximum on a screen at a distance $D$ is given by $W = D(2	heta) = \frac{2D\lambda}{a}$.

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