Question

Light from a sodium lamp $\left(\lambda =589nm\right)$illuminates a narrow slit and is observed on a screen $75cm$behind the slit. The distance between the first and third dark fringes is $7.5mm$. What is the width (in $mm$) of the slit?

Short answer

The slit's width of is $0.12mm.$

Step-by-step solution

Step: 1 Width of slit:

The light casts a shadow when the slit widths are bigger than the wavelength of the sunshine. Light diffraction occurs when the slit widths are small, and also the light waves overlap on the screen. As a result, the sunshine intensity rises because the slit width grows.

Step: 2 Equating part:

In one slit, the location of $p$th darkest fringes is

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

Calculating the difference by

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

Step: 3 Slit's width value:

The width of slit by

$$\begin{gathered} a=\frac{\mathrm{\Delta }p\lambda L}{\mathrm{\Delta }{y}_p} \\ a=\frac{(3-1)\times 5.89\times {10}^{-7}\times 0.75}{7.5\times {10}^{-3}} \\ a=1.178\times {10}^{-4}\mathrm{m} \\ a=0.12\mathrm{mm}. \end{gathered}$$