# Maximum frequency if there is to be one diffraction order

• thomas19981
It's just an estimate, so I'm not sure if the extra factor of 2 is necessary, but it does seem like a reasonable thing to do (to me).In summary, the naval towed-array sonar consists of a line of 80 equally spaced transducers, towed behind a ship for a total length of 120m, and can be electronically steered with an adjustable phase delay. The sonar beam's angular width can be estimated using the equation 2arcsin(1520*80/120f), and the angle of the beam can be determined using the equation arcsin((n*lambda+dsin(delta phi))/d). To find the maximum frequency for 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.

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.

thomas19981
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.)

Last edited:
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?

thomas19981 said:
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.

Last edited:
thomas19981

## 1. What is maximum frequency in relation to diffraction?

The maximum frequency in diffraction refers to the highest frequency of light that can be diffracted by a given diffraction grating. It is determined by the spacing of the grating and the angle of incidence of the light.

## 2. How is maximum frequency calculated for diffraction?

The maximum frequency for diffraction can be calculated using the formula: fmax = n/d, where n is the order of the diffraction and d is the spacing of the grating. This formula is based on the grating equation: nλ = d(sinθi + sinθr), where n is the order, λ is the wavelength of light, θi is the angle of incidence, and θr is the angle of diffraction.

## 3. What happens if the maximum frequency is exceeded in diffraction?

If the maximum frequency is exceeded in diffraction, higher diffraction orders will occur. This means that light of higher wavelengths will also be diffracted, resulting in a broader spectrum. However, this can also lead to overlapping of diffraction orders and a decrease in the overall quality of the diffraction pattern.

## 4. Can the maximum frequency be adjusted in diffraction?

Yes, the maximum frequency in diffraction can be adjusted by changing the spacing of the grating or the angle of incidence of the light. A smaller spacing or a smaller angle of incidence will result in a higher maximum frequency, while a larger spacing or angle of incidence will result in a lower maximum frequency.

## 5. What factors affect the maximum frequency in diffraction?

The maximum frequency in diffraction is affected by the spacing of the grating, the angle of incidence of the light, and the wavelength of the incident light. It is also influenced by the material and thickness of the grating, as well as any optical aberrations present in the system.

• Introductory Physics Homework Help
Replies
5
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
8
Views
2K
• Introductory Physics Homework Help
Replies
4
Views
3K
• Introductory Physics Homework Help
Replies
10
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
4K
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
3
Views
5K
• Optics
Replies
20
Views
2K