Interference: Factors that reduce the widths of primary maxima

In summary: The ## N ## is the number of slits, and ## I_o ## is the interference intensity.) So the width of the primary maxima is ## \Delta \theta = \frac{\lambda}{Nd} ##. Doubling ## N ## doubles the intensity, so the width decreases by a factor of 4.
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Homework Statement
An interference pattern due to a number of slits is incident on a screen. You are able to alter the number of slits, the spacing of the slits, and the wavelength of the light. Order the following adjustments according to which will most effectively reduce the widths of the primary maxima.
A: Double the number of slits
B: Double the wavelength of the incident light
C: Double the frequency of the incident light
D: Double the slit spacing
Relevant Equations
y_m = (m*lambda*D)/(d)
The ranked order from most to least effectively reducing the widths of the primary maxima is: A>D=C>B. I know that doubling the wavelength of incident light will double the width (y) of the fringe. This means that doubling the frequency of incident light will reduce the fringe width (y) by half. Doubling the slit spacing (d) should also reduce the fringe width (y) by half. I don't how much doubling the number of slits will cause the the fringe width to decrease. Is there any relationship between the number of slits and fringe width?
Thank you.
 
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The way I know how to do this kind of calculation is to use the formula for the intensity of ## N ## equally spaced slits: ## I=I_o \frac{\sin^2(\frac{N \phi}{2})}{\sin^2(\frac{\phi}{2})} ## where ## \phi=\frac{2 \pi d \sin{\theta}}{\lambda} ##.
The width ## \Delta \theta ## is computed as follows: Primary maxima occur when the denominator is zero, along with the numerator. That means ## N \phi/2=Nm \pi ##. Meanwhile, the peak has a width determined by the location of the first adjacent zero in the numerator: ## N \phi'/2=Nm \pi+\pi ##. The result is ## N(\phi'-\phi)/2=\pi ## with ## \phi'-\phi \approx \frac{2 \pi d \Delta \theta}{\lambda} ##. With a little algebra, you get ## \Delta \theta=\frac{\lambda}{Nd} ##.
 
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The above formula ## \Delta \theta \approx \frac{\lambda}{Nd} ## would suggest that (A=C=D)>B would be the correct answer. ## \\ ##Giving it a little extra thought, doubling ## N ## will increase the peak intensity by a factor of 4, as well as reducing the width by one half, so the peaks will appear much sharper, and that may be why they selected the answer as A>C=D>B. That could also be why they used the word "effectively" when they phrased the question.:welcome:
 
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Thank. I apologize for not being clear in initial post. I thought the ranking order was A>C=D>B. This is not the answer key. Can you give a more detailed derivation of $$\Delta{\theta} = \frac{\lambda}{Nd}$$
 
  • #5
My derivation is fairly detailed. One of my areas of specialization is diffraction grating type spectroscopy=what I showed is how this calculation is done starting with what is a very well-known formula for the interference intensity from a grating with ## N ## slits of spacing ## d ##. I think that formula is derived in most books that cover diffraction and interference theory in any detail.
The formula is a little tricky to use, because the denominator goes to zero in places=at the primary maxima=at those locations, you take the limit of the ratio of numerator and denominator, and get the result that ## I=N^2 I_o ##.
 
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FAQ: Interference: Factors that reduce the widths of primary maxima

What is interference?

Interference is a phenomenon that occurs when two or more waves overlap and interact with each other. This interaction can result in constructive or destructive interference, which affects the amplitude and intensity of the resulting wave.

What are primary maxima?

Primary maxima are the points of maximum intensity in an interference pattern. These points occur when the waves are in phase and their amplitudes add together, resulting in a larger amplitude.

What factors can reduce the widths of primary maxima?

The widths of primary maxima can be reduced by various factors, including the wavelength of the waves, the distance between the sources, and the angle of incidence. Additionally, the use of a narrow slit or a single source can also result in narrower primary maxima.

How does the wavelength affect the widths of primary maxima?

The wavelength of the waves plays a significant role in determining the widths of primary maxima. As the wavelength decreases, the widths of the maxima also decrease, resulting in a more concentrated interference pattern.

What is the relationship between the distance between sources and the widths of primary maxima?

The distance between sources is inversely proportional to the widths of primary maxima. This means that as the distance between sources increases, the widths of the maxima decrease, resulting in a more concentrated interference pattern.

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