Maximum frequency if there is to be one diffraction order

[thomas19981]

Homework Statement

A naval towed-array sonar comprises a line of $80$ transducers, equally spaced over a total length of $120 m$, that is towed behind a ship so that it lies in a straight line just below the surface of the water. An adjustable phase delay can be introduced electronically for each transducer, allowing the sonar beam to be steered without physically moving the array. The speed of sound in salt water may be taken to be around $1520 ms^{-1}$.
If the transducers are used in phase at a constant frequency $f$, estimate the angular width of the (zeroth order) sonar beam.
A phase delay $\delta \phi$ is now introduced between successive transducers. Determine how the angle $\theta$ through which the beam is steered depends upon $\delta \phi$
Find the maximum frequency that may be used if only one diffraction order is ever to be present as the beam is scanned from $\theta = -90º$ to $\theta = 90º$.

Homework Equations

$n\lambda=dsin(\theta)$
$v=f\lambda$
$dsin(\theta)-dsin(\delta \phi)=n\lambda$

The Attempt at a Solution

So for the first part I used $n\lambda=dsin(\theta)$ and set $n=1$. $\theta=arcsin(\frac{n\lambda}{d})$ so the angular width would be $2\theta=2arcsin(\frac{n\lambda}{d})$. Subbing in the values given and using $v=f\lambda$ gives $2arcsin(\frac{1520*80}{120f})$. Is this the right approach.

For the second part I considered it as plane waves incident on a double slit which was at an angle $\delta \phi$. I know that this is not a double slit but I guessed the fact that the formula for maxima for a double slit and a diffraction grating are the same it would be ok. Anyways this eventually came to be one of the "relevant equations": $dsin(\theta)-dsin(\delta \phi)=n\lambda$. So I just rearranged this for $\theta$ which came to $\theta=arcsin(\frac{n\lambda+dsin(\delta \phi)}{d})$.

The third part is where I get stuck. Any help would be very much appreciated.

 

[Charles Link]

For the width of the zeroth order beam, you need the equation $I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)}$ where $\phi=\frac{2 \pi}{\lambda} d \sin(\theta)$ and $N=80$. The width is estimated by the angle for which the numerator encounters its first zero from the angle where the maximum occurs at. The primary maxima occur when the denominator equals zero. The numerator is also equal to zero, and in the limit, $m \lambda=d \sin(\theta)$, $I=N^2 I_o$ at the primary maxima.

 

[Charles Link]

For the second part I think they are wanting basically the wavelength for which the $m=1$ and/or $m=-1$ start to occur. (They are of course counting the $m=0$ as a maximum). That means we must have $(1)(\lambda)=d[\sin(\theta)+ \sin(\delta)]$. Just as a quick observation=I don't know if it's correct, I haven't analyzed it in sufficient detail=I'll let you determine that=if $\lambda>2 d$, this $m=1$ can not occur, because $|\sin(\theta)| \leq 1$. $\\$ Editing: You correctly determined for this problem that the phase (between adjacent sources) is $\phi=\frac{2 \pi}{\lambda} d [\sin(\theta) \pm \sin(\delta)]$. (Either + or - would be correct=it's a matter of geometry which one you choose.)

 

[thomas19981]

For the width of the zeroth order beam, you need the equation $I(\theta)=I_o \frac{\sin^2(N \phi/2)}{\sin^2(\phi/2)}$ where $\phi=\frac{2 \pi}{\lambda} d \sin(\theta)$ and $N=80$. The width is estimated by the angle for which the numerator encounters its first zero from the angle where the maximum occurs at. The primary maxima occur when the denominator equals zero. The numerator is also equal to zero, and in the limit, $m \lambda=d \sin(\theta)$, $I=N^2 I_o$ at the primary maxima.

When I solve for $\theta$ (for the first part) would I double it to get the width?

 

[Charles Link]

When I solve for $\theta$ (for the first part) would I double it to get the width?

It's really only an estimate. The FWHM (full width at half-max) is approximately the position of the first zero from center. Usually I think that just the first zero from center is used to estimate the FWHM. You can google Rayleigh criteriion if you want. I am presently retired, but I did do a fair amount of (diffraction grating type) spectroscopy=let me google that, and see what they say...$\\$ Editing: The Rayleigh criterion is for resolving two lines in a spectrometer of slightly different wavelengths... To answer your question, I think either answer would be correct, but your extra factor of 2 might be a better estimate. There a couple of somewhat significant secondary maxima on both sides of the first zeros from the primary maximum, so that your extra factor of 2 would certainly give a reasonable estimate on the angular spread of the beam.