91Ó°ÊÓ

• Width of slit, $a=0.750mm=0.750×{10}^{-3} m$
• Distance between screen and slit, R = 2.00 m
• Distance of first minima from central maxima, $y=1.35mm=1.35×{10}^{-3}m$

Diffraction is a process by which a beam of light spreads when passing through a narrow passage or across the edge of an obstacle accompanied by interference between the waveforms.

By using the equation:

$$\begin{gathered} \sin \mathrm{θ}=\frac{m\mathrm{λ}}{a} \\ \mathrm{λ}=\frac{a\sin \mathrm{θ}}{m} \end{gathered}$$

Here, a is the slit width, $\mathrm{λ}$is the wavelength of light used and is the diffraction angle.

For first minima, m = 1

$\mathrm{λ}=a\sin \mathrm{θ}$

To obtain the value of θ

$$\begin{gathered} tan\mathrm{θ}=\frac{y}{R} \\ \mathrm{θ}=ta{n}^{-1}\frac{y}{R} \\ \mathrm{θ}=ta{n}^{-1}\left(\frac{1.35×{10}^{-3} m}{2.0m}\right) \\ \mathrm{θ}=0.0387° \end{gathered}$$

Therefore, the wavelength of source is given by

$$\begin{gathered} \mathrm{λ}=a\sin \mathrm{θ} \\ =0.750×\sin \left(0.0387°\right) \\ =5.06×{10}^{-7}m \\ =506nm \end{gathered}$$

Therefore, the wavelength of the light source is 506 nm.