Question

Light of$630\mathrm{nm}$wavelength illuminates a single slit of width $0.15\mathrm{mm}$. FIGURE $\mathrm{EX}33.17$shows the intensity pattern seen on a screen behind the slit. What is the distance to the screen$?$

Short answer

Distance to the screen is$1.19\mathrm{m}$.

Step-by-step solution

Formula for length

Between the primary order minima on each side of the screen, the centre maximum stretches up to the distance between them.

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

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

So,

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

localid="1649234782892" $\mathrm{L}=\frac{\mathrm{wa}}{2\mathrm{\lambda }}$

Calculation for distance of the screen

Width is,

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

The screen's distance,

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

$=\frac{\left(1\times {10}^{-2}\mathrm{m}\right)\times \left(0.15\times {10}^{-3}\mathrm{m}\right)}{2\times 630\times {10}^{-9}\mathrm{m}}$

$\mathrm{L}=1.19\mathrm{m}$