Question

A $0.50-\mathrm{mm}$-wide slit is illuminated by light of wavelength $500\mathrm{nm}$. 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$4\mathrm{mm}$.

Step-by-step solution

Formula for width

The focus maximum is equal to the distance between the first order minima along both sides of the screen, yet it is positioned between the minima.

$2\mathrm{Da}$is the maximum width of the centre.

$2\mathrm{\theta }=2\mathrm{\lambda a}$is the centre maximum's angular width.

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

Calculation for width

The width of the median maximum in the scientific experiment is

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

This width will be in our case

localid="1649240628379" $\mathrm{w}=\frac{2\times 5\times {10}^{-7}\times 2}{5\times {10}^{-4}}$

$=4\times {10}^{-3}\mathrm{m}$

$=4\mathrm{mm}$