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A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of ? = 561 nm. An interference pattern is observed on a screen 122

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asked 2021-03-26
A pair of slits, separated by 0.150 mm, is illuminated by light having a wavelength of ? = 561 nm. An interference pattern is observed on a screen 122 cm from the slits. Consider a point on the screen located at y = 2.00 cm from the central maximum of this pattern.
(a) What is the path difference ? for the two slits at the location y?
(b) Express this path difference in terms of the wavelength.

Answers (1)

2021-03-28
our formula for double slit interference is:
\(\displaystyle{d}{\sin{{\left({t}\right)}}}={m}{\left({w}\right)}\)
where d is the width of the slits
where t is the angle from the slits
where m is the corresponding fringe from the central fringe
where w is the wavelength
using trig, we can find \(\displaystyle{\sin{{\left({t}\right)}}}\) to be:
\(\displaystyle{\sin{{\left({t}\right)}}}={\frac{{{y}}}{{{L}}}}\)
where y is the distance from central fringe to corresponding fringe
where L is the distance from slit to screen
our formula now becomes:
\(\displaystyle{d}{\left(\frac{{y}}{{L}}\right)}={m}{\left({w}\right)}\)
where both sides represent the path difference:
path difference=d(y/L)
path difference = .150e-3(2e-2/122e-2)
path difference = 2.459e-6 m
In terms of the wavelength:
2.459e-6/561e-9 = 4.4w
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