(a) The minimum uncertainty in the vertical component of the momentum of each photon in the beam is [tex]5.33 \times 10^{-30}\;\rm kg.m/s[/tex].
(b) The width of the central diffraction maximum observed on the screen is [tex]2.48 \times 10^{-4} \;\rm m[/tex].
Given data:
The wavelength of laser light beam is, [tex]\lambda = 585\;\rm nm =585 \times 10^{-9} \;\rm m[/tex].
The width of narrow slit is, [tex]a = 0.0620 \;\rm mm =0.0620 \times 10^{-3} \;\rm m[/tex].
The distance between the slit and screen is, [tex]D = 2.00 \;\rm m[/tex].
(a)
The minimum uncertainty in the vertical component of the momentum of each photon is given as,
[tex]a \times \Delta P =\dfrac{h}{2}[/tex]
Here, h is the Planck's constant.
[tex](0.0620 \times 10^{-3}) \times \Delta P =\dfrac{6.62 \times 10^{-34}}{2}\\\\\Delta P = \dfrac{6.62 \times 10^{-34}}{2 \times (0.0620 \times 10^{-3})}\\\\\Delta P = 5.33 \times 10^{-30}\;\rm kg.m/s[/tex]
Thus, we can conclude that the minimum uncertainty in the vertical component of the momentum of each photon in the beam is [tex]5.33 \times 10^{-30}\;\rm kg.m/s[/tex].
(b)
The width of the central diffraction maximum that is observed on the screen
is obtained by the expression,
[tex]w' = \dfrac{D \times h}{\Delta P}[/tex]
Solving as,
[tex]w' = \dfrac{2.00 \times (6.62 \times 10^{-34})}{5.33 \times 10^{-30}}\\\\w' = 2.48 \times 10^{-4} \;\rm m[/tex]
Thus, we can conclude that the width of the central diffraction maximum observed on the screen is [tex]2.48 \times 10^{-4} \;\rm m[/tex].
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