# 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

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.
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Faiza Fuller
our formula for double slit interference is:
$d\mathrm{sin}\left(t\right)=m\left(w\right)$
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 $\mathrm{sin}\left(t\right)$ to be:
$\mathrm{sin}\left(t\right)=\frac{y}{L}$
where y is the distance from central fringe to corresponding fringe
where L is the distance from slit to screen
our formula now becomes:
$d\left(\frac{y}{L}\right)=m\left(w\right)$
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