Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04μm apart, and in line with an observer, so that one source is 2.04μm farther from the observer than the other.

Part A

For what visible wavelengths (400 to 700 nm) will the observer see the brightest light, owing to constructive interference?

Part B

What would your answers to part (a) be if the two sources were not in line with the observer, but were still arranged so that one source is 2.04μm farther away from the observer than the other?

Part C

For what visible wavelengths will there be destructive interference at the location of the observer?

Question

Physics

Ireland

5 January, 22:46

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04μm apart, and in line with an observer, so that one source is 2.04μm farther from the observer than the other.

Part A

For what visible wavelengths (400 to 700 nm) will the observer see the brightest light, owing to constructive interference?

Part B

What would your answers to part (a) be if the two sources were not in line with the observer, but were still arranged so that one source is 2.04μm farther away from the observer than the other?

Part C

For what visible wavelengths will there be destructive interference at the location of the observer?

+5

Answers (1)

Know the Answer?

Jamison Michael · Answer 1

Path difference for the observer = 2.04μm

For constructive interference

path difference = n λ

2.04μm = 5 x. 408 or, 4 x. 51 or 3 x. 68μm

For visible wavelengths like. 408,.51 or. 68 μm (400 to 700 nm) will the observer see the brightest light due to constructive interference.

B) If the two sources were not in line with the observer, but were still arranged so that one source is 2.04μm farther away from the observer than the other, in that case also, interference pattern will be visible.

c)

For destructive interference

2.04 μm = (2n+1) λ / 2

2.04μm = 3.5 x. 58μm

2.04 μm = 4.5 x. 45μm

So for. 58μm and. 45μm, destructive interference will take place