A horizontal beam of laser light of wavelength 585 nm passes through a narrow slit that has width 0.0620 mm. The intensity of the light is measured on a vertical screen that is 2.00 m from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

A. 8.51x10^-31 kgm/s

B. 1.51mm

Explanation:

We have a∆Py >= h/2

∆Py = uncertainty

a = width

We calculate ∆Py =

1.055x10^-34/2x(0.0620x10^-3)

= 8.51x10^-31 kg m/s

This is the maximum value of uncertainty

B.

h/lambda

Lambda = 585

h = 6.626x10^-34

= 6.626x10^-34 / 585x10^-9

= 1.13x10^-27

From our answer in part a, we solve for the width

8.51x10^-31 / 1.13x10^-27

= 7.53x10^-4

Width = 2 * 7.53x10^-4

= 1.55mm

Please check attachment for the solution I provided

c, 1a hope this

double the distance = divide strength by 4