The intensity on the screen at a certain point in a double slit interference pat

ID: 1648391 • Letter: T

Question

The intensity on the screen at a certain point in a double slit interference pattern is 60% of the maximum value.

a) What minimum phase difference (in radians) between sources produces this result?

b) Express this phase difference as a path difference for 600 nm light.

c) If the slits are separated by 0.18 mm and the screen is 80.0 cm away, calculate the distance y above the central maximum for point P.

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A chemical film (n=1.25) lying on top of oil (n=1.45) is illuminated by white light at normal incidence.

d) The film is 140 nm thick. Find the color of the light in the spectrum least reflected.

part a:

intensity as a function of phase difference is given as:

I=I0*cos^2(phi/2)

where phi=phase difference

given I=0.6*I0

==>0.6=cos^2(phi/2)

==>phi=1.3694 rad

part b:

phase difference=2*pi*path difference/wavelength

==>path difference=1.3694*600 nm/(2*pi)=130.77 nm

part c:

as we know, path difference=slit width*distance on the screen from central maximum/distance of the screen from the slits

==>130.77*10^(-9)=0.18*10^(-3)*y/0.8

==>y=0.58121 mm

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