91Ó°ÊÓ

It is given that,

• Wavelength of light source, $\mathrm{λ}=546nm=546×{10}^{-9} m$
• Distance between screen and slit, R = 60 cm = 0.6 m
• Distance of first minima from central maxima,$y=8.65mm=8.65×{10}^{-3} m$

The turning around of light rays from the edges of the object and entering the shadow region creating an interference pattern is known as diffraction.

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$\mathrm{θ}$is the diffraction angle.

For first minima, m = 1

$\mathrm{λ}=a\sin \mathrm{θ}$

To obtain the value of θ

$$\begin{gathered} tan\mathrm{θ}=\frac{y}{R} \\ \mathrm{θ}=ta{n}^{-1}\left(\frac{y}{R}\right) \\ \mathrm{θ}=ta{n}^{-1}\left(\frac{8.65×{10}^{-3} m}{0.6m}\right) \\ \mathrm{θ}=0.826° \end{gathered}$$

Therefore, the wavelength of source is given by

$$\begin{gathered} a=\frac{\mathrm{λ}}{\sin \mathrm{θ}} \\ =\frac{546×{10}^{-9}m}{\sin \left(0.826°\right)} \\ =3.79×{10}^{-5}m \\ =3.79\mathrm{μ}m \end{gathered}$$

Therefore, the width of the slit is $3.79\mathrm{μ}m$.