Inferring Shape of Phasors in Multi-Slit Diffraction

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Phasors in multiple-slit diffraction differ significantly from those in single-slit diffraction. In single-slit analysis, phasors are associated with each point in the aperture and align only in the central direction, leading to minima when they form a closed polygon. Conversely, in multiple-slit scenarios, each slit corresponds to a single phasor, and maxima occur when these phasors are aligned. The phase angle between adjacent sources is crucial for understanding the interference pattern, as it determines the alignment of phasors. Overall, careful comparison of single-slit and double-slit simulations enhances understanding of these concepts.
hidemi
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Homework Statement
At a bright diffraction line phasors associated with waves from the slits of a multiple-slit barrier:
A. are aligned
B. form a closed polygon
C. form a polygon with several sides missing
D. are parallel but adjacent phasors point in opposite directions
E. form the arc of a circle

The correct answer is A
Relevant Equations
d * sin(theta) = m * lambda
I know that phasors of a single-slit diffraction form a closed polygon or circle, but how could we infer the shape when phasors generated by slits of a multiple-slit barrier?
 
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The ## m \lambda=d \sin{\theta} ## for constructive interference is sort of on the right track, but what you are needing is the phase angle between phasors from adjacent sources a distance ## d ## apart: ## \phi=(\frac{2 \pi}{\lambda}) d \sin{\theta} ##. Using the first expression, (since you are told that the sources constructively interfere), what can you say about ## \phi ##? Will the phasors be aligned?
 
hidemi said:
I know that phasors of a single-slit diffraction form a closed polygon or circle ...
Closed polygon's give minima (zero intensity), not maxima.

Think of adding phasors in the same way as adding vectors. The resultant is zero only when the vectors form a closed polygon.

There is an important difference between a single-slit and multiple-slits when using phasors.

For a single-slit analyis, each point in the aperture has a phasor. The phasors are aligned in only one situation - for the central direction.

For a multiple-slit analysis we associate each slit with a single phasor. A maximum is produced whenever the phasors are aligned (unlike a single-slit).

This video gives quite a good insight. You need to compare the single-slit and double-slit simulations carefully.
 
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Steve4Physics said:
Closed polygon's give minima (zero intensity), not maxima.

Think of adding phasors in the same way as adding vectors. The resultant is zero only when the vectors form a closed polygon.

There is an important difference between a single-slit and multiple-slits when using phasors.

For a single-slit analyis, each point in the aperture has a phasor. The phasors are aligned in only one situation - for the central direction.

For a multiple-slit analysis we associate each slit with a single phasor. A maximum is produced whenever the phasors are aligned (unlike a single-slit).

This video gives quite a good insight. You need to compare the single-slit and double-slit simulations carefully.

Thank you! I got it
 
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My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...

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