91Ó°ÊÓ

Here given the wavelength of the laser beam$\mathrm{λ}=585nm=585×{10}^{-9}m$.

Since the width of the narrow slit is perpendicular to the laser light, say laser light moving in the x-direction, then the slit width will be along the y-axis. Hence the width of the slit$a=∆y=0.062mm=0.062×{10}^{-3}m$

Now from the Heisenberg uncertainty principle for position and momentum is

$∆y∆P_y=\frac{h\text{'}}{2}$ (1)

Where$∆y$uncertainty in vertical coordinate y, $∆p_y$is uncertainty in the corresponding momentum component and $h\text{'}=\frac{h}{2\mathrm{π}}=1.054×{10}^{-34}J.s$is Planck's constant divided by$2\mathrm{π}$

Hence from equation (1), the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit will be

$$\begin{gathered} ∆p_y=\frac{h\text{'}}{2∆y} \\ =\frac{1.05×{10}^{-34}Js}{2×0.062×{10}^{-3}m} \\ =8.5×{10}^{-31}kgms \end{gathered}$$

As we know the angle of diffraction corresponding to the first minima is given by

$$\begin{gathered} \mathrm{θ}=\frac{∆p_y}{p_x} \\ =\frac{∆p_y\mathrm{λ}}{h} \\ =\frac{8.5×{10}^{-31}kgm{s}^{-1}×585×{10}^{-9}m}{6.626×{10}^{-34}Js} \\ =7.5×{10}^{-4}rad \end{gathered}$$

Thus the total width of the central diffraction maximum is approximately given by

$W=2d\tan \mathrm{θ}$ (3)

Where d =2 m is the distance of the screen from the slit

So from equations (2) and (3), we get

$$\begin{gathered} w=2×2m×\tan \left(7.5×{10}^{-4}rad\right) \\ =2×2m×\tan \left(0.04297°\right) \\ =3×{10}^{-3}m \\ =3mm \end{gathered}$$