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Question

I cannot seem to solve this problem for my life. To begin, is this an interference problem or double-slit problem? Anyway.. more importantly is the procedure to get the answer. Here'e the question:

Two line sources transmit microwaves of wavelength 3 cm in phase. How far apart should they be in order to produce a 10 degree angular seperation between the first and second order bright fringes on one side of the central peak?

Thanks a bunch if you took the time to answer!

Answer 1

This is a little bit unclear, but I will answer the question I think is being asked.

I assume we are talking about two sources, each in the shape of a line of infinite length, and parallel to each other. In that case, suppose they are each parallel to the z-axis, then we can look at what is going on in the x-y plane.

Along the line in the x-y plane which is equidistant from the intersection of the two lines with the plane (two points), the two phases (one from each source) are the same, because each point on that line is equidistant from the two sources. This is the "central peak": at any distance, the total intensity of the signal is highest here, because the two signals will always be exactly in-phase.

If you now create a second line, at angle phi from the equidistant line, and passing through the point directly between the two intersection points, the distance from a point on that line to one of the sources will be different than the distance to the other source. If the distance from that point to the sources is very large, then there is still a difference in distance between the two. Since the angle between the point of interest and the "central peak" line is phi, the difference in distances is proportional to the distance between the sources (d). In fact, it should be d sin(phi). (It's hard to describe this without a picture.)

So since the difference between the phases is
((2 pi)/wavelength)*(difference in distances)
= (2 pi d sin(phi))/wavelength
= (2 pi d sin(phi))/(3 cm)

the first-order fringe is when this phase difference = 2 pi, and the second-order fringe is when it = 4 pi.

Therefore:
- first-order:
2 pi = (2 pi) (d sin(phi-1))/(3 cm)
d sin(phi-1) = 3 cm

- second-order:
4 pi = (2 pi) (d sin(phi-2))/(3 cm)

d sin(phi-2) = 6 cm

How to solve this? Note that phi-2 = phi-1 + X, where X = 10 degrees.

Thus,
d sin(phi-1) = 3 cm
d sin(phi-1 + X) = 6 cm = 2 d sin(phi-1)
sin(phi-1 + X) = 2 sin(phi-1)
sin(phi-1) cos(X) + cos(phi-1) sin(X) = 2 sin(phi-1)
cos(phi-1) sin(X) = (2-cos(X))sin(phi-1)
sin(X)/(2-cos(X)) = tan(phi-1)

OK:
a) Since X = 10 degrees, phi-1 is the angle at which
tan(phi-1) = sin(10 degrees)/(2 - cos(10 degrees))

When I work this out in Excel, I find that:
sin(10 degrees) = 0.1736
cos(10 degrees) = 0.9848
tan(phi-1) = 0.1710
phi-1 = 0.16941 rad = 9.7 degrees

b) Once you've found phi-1, then
d = (3 cm)/sin(phi-1)
= (3 cm)/0.168601
= 17.7935 cm

Answer 2

It is both. Calm down and draw it on paper. You are determined_ladii, so I'm sure you'll get it. Note that the two waves will be at max interference when the distances from the slits to the point on a screen differ by an integer (N)multiple of wavelengths. First order means the N=1, second order means N=2. Draw screen distance L from the slits of separation S. Draw a triangle with verticies at slits and N=1 and another triangle with a vertex a N=2. Solve for S. Take L approaches infinity (otherwise question is vague).

Answer 3

Band width of the first and second bright fringe is (D/d) λ = (D/d) 0.03.

If θ is the angular separation, (D/d) 0.03 = D θ

d = 0.03 / θ

10 degree = 0.1745 radian.

d = 0.03 / 0.1745 = 0.17 m