Calculating slit separation for sound two-slit interference?

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The discussion revolves around calculating the slit separation for a two-slit interference setup to achieve the same interference angles for sound as for red light. The original setup has a slit separation of 0.30 mm for red light at a frequency of 4.9×10^14 Hz. The analogous setup for room-temperature sound at a frequency of 262 Hz requires a slit separation of 640 mm. The key realization is that sound travels at a different speed than light, which affects the calculations. The solution was confirmed after clarifying the relationship between frequency, wavelength, and velocity in the context of sound.
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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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