Calculating slit separation for sound two-slit interference?

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Homework Help Overview

The problem involves a two-slit interference setup where the original poster seeks to determine the necessary slit separation for sound waves to produce interference fringes at the same angles as those produced by red light. The original setup has a known slit separation and frequency for light, while the analogous setup involves sound at a significantly lower frequency.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between wavelength and frequency, questioning how to apply the interference formula given the differing velocities of light and sound. There is an exploration of whether the velocities can be considered equal and how that affects the calculations.

Discussion Status

The discussion has seen participants questioning assumptions about wave velocities and their implications for the problem. Some guidance has been offered regarding the correct application of the wave equation, indicating a productive direction in the conversation.

Contextual Notes

There is a notable distinction between the speed of light and the speed of sound, which has been acknowledged in the discussion. The original poster's confusion regarding the application of the formula has been addressed, but the exact resolution remains unclear.

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Homework Statement



A two-slit interference set-up with slit separation d =0.30 mm produces interference fringes at a particular set of angles THETAm (where m = 0, 1, 2,) for red light of frequency f = 4.9×10^14 hz.

If one wishes to construct an analogous two-slit interference set-up that produces interference fringes at the same set of angles THETAm for room-temperature sound of middle-C frequency f = 262 hz, what should the slit separation distance be for this analogous set-up?
this particular problems answer is 640 but i can't seem to arrive at that

Homework Equations


sin(theta)= m(wavelength)/d

The Attempt at a Solution


i believe sin(theta) is equalin both, so i don't see why it wouldn't come down to wavelenth/d is proprotional to wavelenth/d, or converting to frequency 1/fd = 1/fd or fd=fd.
 
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You know \lambda = v/f, but they both have different velocities so you can't use fd = fd.
 
arent their velocities equal at c, being the speed of light?
 
room-temperature sound of middle-C frequency f = 262 hz

Sound doesn't travel at the speed of light. At least not where I live at.
 
oh that is quite true, shouldn't have overlooked that, so i can now say the formula is
c/fd = 343/fd?

edit: yes i can, thanks a lot nickjer, i got it now.
 

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