Why Does the mth Interference Maximum Vanish in Single Slit Diffraction?

[slyman]

Two narrow slits of width a are separated by center-to-center distance d. Suppose that the ratio of d to a is an integer d/a=m.

Show that in the diffraction patterns produced by this arrangement of slits, the mth interference maximum (corresponding to dsin(theta)=m$$\lambda$$) is suppressed because of coincidence with a diffraction minimum. Show that this is also true for the 2mth, 3mth, etc., interference maxima.

My attempt dsin(theta)=m$$\lambda$$ indicates the conditions for interference maxima, where d is the distance between the two slits. We know that diffraction minimum occurs when a*sin(theta)=$$\lambda$$, where a is the slit width. Divide by both equations and we can determine which interference maximum coincides with the first diffraction minimum: d/a=mI don't really understand what the question wants. Could someone please help me out? Thanks in advance.

 

[Gokul43201]

Please read the posting guidelines and use the template provided. We can not help you unless you first show some effort yourself.

 

[slyman]

My bad. It's just that this question pisses me off.

 

[Gokul43201]

Okay, you've almost answered the first part of the question, but you've gone about it backwards.

Let $d\cdot sin\theta = m \lambda$ and $d/a=m$, then you can show that $a \cdot sin \theta = (d/m)sin \theta =...$..which answers the first part of the question.

For the second part, how would you proceed?