Question

Light of wavelength $\mathbf{500}\mathbf{}\mathit{n}\mathit{m}$ diffracts through a slit of width$\mathbf{2}\mathit{\mu }\mathit{m}$ and onto a screen that is $\mathbf{2}\mathit{m}$away. On the screen, what is the distance between the center of the diffraction pattern and the third diffraction minimum?

Short answer

The distance between the center of the diffraction pattern and the third diffraction minimum is$1.5mm$ .

Step-by-step solution

Given data:

The wavelength of incident light $\lambda =500n\mathrm{m}$:

Slit width: $d=2mm$

Distance of screen from slit: $D=2m$

Diffraction from a single slit:

The distance of the ${\mathit{m}}_{\mathbf{t}\mathbf{h}}$ order single slit diffraction minima from thecentral maxima for slit width $\mathit{d}$ and screen distance $\mathit{D}$ is

$\mathit{x}\mathbf{=}\frac{\mathbf{m}\mathbf{\lambda }\mathbf{D}}{\mathbf{d}}$ .....(1)

Here,$\mathit{\lambda }$ is the wavelength of the incident light.

Determining the distance of the third minimal from central maxima

Calculate the distance of the third minima from the central maxima by substituting known values into equation (1).

$\begin{array}{l}x=\frac{3\times 500\times {10}^{-9}m\times 2m}{2\times {10}^{-3}m}\\ =0.0015m\\ =1.5mm\end{array}$

Thus, the required distance is $1.5mm$.