Question

In a single-slit experiment, the slit width is $200$ times the wavelength of the light. What is the width $(\mathrm{in}\mathrm{mm})$of the central maximum on a screen $2.0\mathrm{m}$ behind the slit$?$

Short answer

Width of the central maximum is$20\mathrm{mm}$.

Step-by-step solution

Formula for width

The centre maximum is equal to the distance between the first order minima on both sides of the screen, and it is placed between the minima.

Width of center maximum is$2\mathrm{Da}$.

Angular width of central maximum is$2\mathrm{\theta }=2\mathrm{\lambda a}$.

So,

$\mathrm{w}=\frac{2\mathrm{\lambda L}}{\mathrm{a}}$

Calculation for width

The breadth of the centre maximum is computed as follows:

$\mathrm{w}=\frac{2\mathrm{\lambda L}}{\mathrm{a}}$

$=2\mathrm{L}\frac{\mathrm{\lambda }}{200\mathrm{\lambda }}$

$=\frac{\mathrm{L}}{100}$

Width is,

$\mathrm{w}=\frac{2}{100}$

$=0.02\mathrm{m}$

$=20\mathrm{mm}$