Solve Double Slit Maxima with 3 Narrow Slits

[frostchaos123]

Homework Statement

3 narrow slits with spacing of d and 3d/2 as in picture, the slits are irradiated from left with a plane of monochromatic light with wavelength 2d/5.

If the bottom slit is covered with a filter that introduces a half phase change, how many
principal maxima will be observed?

Homework Equations

2 slit constructive interference: d sin(theta)= m * wavelength
destructive interference: d sin(theta) = (m + 0.5) * wavelength

The Attempt at a Solution

The top slit has constructive intereference of d * sin(theta) = m * (2d/5), which gives
sin(theta) = 0, 2/5, 4/5 for m = 0,1,2,...

The bottom slit has destructive intereference of 3d/2 * sin(theta) = (m+1/2) * (2d/5), which gives
sin(theta) = 2/15, 6/15, 10/15, ... for m = 0,1,2,...

My question is that the answer given is that sin(theta) = 6/15 will be the maxima, since it occurs at for both top and bottom slit, but why is it so?

Shouldn't the destructive interference of the bottom slit cancel out the constructive interference at sin(theta) = 6/15, making it not the maximum?

Thanks.

 

[vela]

Homework Statement

3 narrow slits with spacing of d and 3d/2 as in picture, the slits are irradiated from left with a plane of monochromatic light with wavelength 2d/5.

If the bottom slit is covered with a filter that introduces a half phase change, how many
principal maxima will be observed?

Homework Equations

2 slit constructive interference: d sin(theta)= m * wavelength
destructive interference: d sin(theta) = (m + 0.5) * wavelength

The Attempt at a Solution

The top slit has constructive intereference of d * sin(theta) = m * (2d/5), which gives
sin(theta) = 0, 2/5, 4/5 for m = 0,1,2,...

The bottom slit has destructive intereference of 3d/2 * sin(theta) = (m+1/2) * (2d/5), which gives
sin(theta) = 2/15, 6/15, 10/15, ... for m = 0,1,2,...

This would be true if no filter were present in front of the bottom slit. The extra phase change the filter introduces cancels part of the phase difference due to the difference in path length, so the waves from the middle and bottom slits constructively interfere when this condition holds.

My question is that the answer given is that sin(theta) = 6/15 will be the maxima, since it occurs at for both top and bottom slit, but why is it so?

Shouldn't the destructive interference of the bottom slit cancel out the constructive interference at sin(theta) = 6/15, making it not the maximum?

Thanks.