A horizontal beam of laser light of wavelength 589 nm passes through a narrow sl

Question

A horizontal beam of laser light of wavelength 589 nm passes through a narrow slit that has width 0.0550 mm. The intensity of the light is measured on a vertical screen that is 2.80 m from the slit. 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? Express your answer with the appropriate units. Use the result of part A to estimate the width of the central diffraction maximum that is observed on the screen. Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

part A:

from heisenberg uncertainty principle:

uncertainty in position*uncertainty in momentum > h/(4*pi)

here uncertainty in psoition=slit width=0.055 mm=55*10^(-6) m

then uncertainty in momentum > 6.626*10^(-34)/(4*pi*55*10^(-6))=9.5869*10^(-31) kg.m/s

part B:

momentum along horizontal direction=h/wavelength

=6.626*10^(-34)/(589*10^(-9))

=1.125*10^(-27) kg.m/s

so angle made by the cental maximum=arctan(uncertainty in momentum along vertical direction / momentum along horizontal direction)

=arctan(9.5869*10^(-31)/(1.125*10^(-27))

=8.5217*10^(-4) rad

then width of central maxima=2*D*tan(theta)

where D=distance of the screen from the slit

=2*2.8*tan(8.5217*10^(-4))

= 0.0047722 m

=4.7722 mm

in two significant figures,

answer=4.8*10^(-3) m

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