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I wonder if someone could help check something for me in order to make sure that I'm not making a stupid mistake with this problem as I've been marked wrong on an undergraduate paper, but I'm almost certain that I'm right. I don't quite have enough courage in my convictions to approach the tutor yet, hence why I'm posting this thread.

## Homework Statement

The question relates to the superposition of waves and destructive and constructive interference in various types of microscopes, and the question as stated is as follows:

"The two light waves below are shifted one quarter out of phase, draw the resultant wave if these are combined"

The paper shows two waves of equal amplitude, frequency, and wavelength, propagated in the same direction, a quarter of a wavelength out of phase.

The feedback I received was that the resultant amplitude should be smaller than the original, as destructive interference occurs, while I drew a wave which was amplified showing constructive interference.

## Homework Equations

The one i'm using is:

Resultant amplitude = 2 x Amplitude (cos [phase shift in radians/2])

I hope that's correct.

## The Attempt at a Solution

I assume I'm correct in stating there are 360 degrees in a wave cycle, so in radians that would be 2 times pi, or approximately 6.283.

If the waves are 1/4 out of phase, the phase shift in radians is 2pi/4 (or pi/2), and so therefore approximately 1.57. Dividing this by 2 we get 0.785, and cos of this (in radians) is 0.707. If we take the amplitude of the original waves as 1, then the resultant amplitude is (2 x 1) x 0.707 = 1.414. The resultant amplitude is therefore larger than the original amplitude.

The formula seems to be correct since at half a wavelength out of phase, which is where I know the resultant amplitude is zero, the answer comes out at zero: (2x1) x cos (pi/2) = 2 x 0 = 0

And when maximum constructive interference occurs, i.e. the waves are in phase, (2x1) x cos (0/2) = (2x1) x 1 = 2

Assuming the equation is correct, you would have to shift a third out of phase before the amplitude matches the original amplitude, and then between a third and half a wavelength out of phase, the interference is destructive down to zero:

(2x1) x cos(2.094/2) = 2 x 0.5 = 1

So anyway, can anyone spot a mistake?

Thanks for any help!