Interference: Factors that reduce the widths of primary maxima

[roya98864]

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.

 

[Charles Link]

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 $\Delta \theta=\frac{\lambda}{Nd}$.

 

[Charles Link]

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.

 

[roya98864]

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 $$\Delta{\theta} = \frac{\lambda}{Nd}$$

 

[Charles Link]

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 $I=N^2 I_o$.