FIGURE $P33.56$shows the light intensity on a screen behind a single slit. The wavelength of the light is$600\mathrm{nm}$and the slit width is $0.15\mathrm{mm}$. What is the distance from the slit to the screen?

Using the dark peripheral condition of diffraction, compute the difference seen between slit and the screen.

The black fringe in a diffraction pattern must meet the following conditions:

$y_p=\frac{p\lambda L}{a}$

$p=1,2,3,......$is an integer,$\lambda$is the light wavelength, $L$is the distance between the slit and the monitor, and $a$is the slit width.

For $L$, rewrite the equation.

$L=\frac{a_p}{p\lambda }$

Convert the slit width measurement from $\mathrm{mm}$to$\mathrm{m}$

$a=0.15\mathrm{mm}$

$=0.15\mathrm{mm}\left(\frac{1\mathrm{m}}{1\times {10}^3\mathrm{mm}}\right)$

role="math" localid="1651344895292" $=0.15\times {10}^{-3}\mathrm{m}$

Convert the frequency of beam's units from $nm$to$m$

$\lambda =600\mathrm{nm}$

$=600\mathrm{nm}\left(\frac{1\mathrm{m}}{1\times {10}^9\mathrm{nm}}\right)$

role="math" localid="1651345038097" $=600\times {10}^{-9}\mathrm{m}$

The distance between the first minimum and the middle bright maximum in the figure is as regards.

$y_p=0.5\mathrm{cm}$

$=0.5\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)$

$=0.5\times {10}^{-2}\mathrm{m}$

Length is,

$L=\frac{a{y}_p}{p\lambda }$

Where,

$\lambda =600nm$

Slit width is $0.15nm$

So,

$L=\frac{\left(0.15\times {10}^{-3}\mathrm{m}\right)\left(0.5\times {10}^{-2}\mathrm{m}\right)}{\left(1\right)\left(600\times {10}^{-9}\mathrm{m}\right)}$

$=1.25\mathrm{m}$