A two-slit interference set-up with slit separation d= 0.10 mm produces interference fringes at

step 1 Okay, so this is kind of a fun problem. It gives you a situation for light interference, two -sl

A two-slit interference set-up with slit separation d= 0.10...

A two-slit interference set-up with slit separation $d=$ 0.10 $\mathrm{mm}$ produces interference fringes at a particular set of angles $\theta_{m}$ (where $m=0,1,2, \ldots )$ for red light of frequency $f=4.6 \times 10^{14} \mathrm{Hz}$ . If one wishes to construct an analogous two-slit interference set-up that produces interference fringes at the same set of angles $\theta_{m}$ for room-temperature sound of midde-C frequency $f_{\mathrm{s}}=262 \mathrm{Hz}$ , what should the slit sepa- ration $d_{\mathrm{S}}$ be for this analogous set-up?

A two-slit interference set-up with slit separation $d=$
0.10 $\mathrm{mm}$ produces interference fringes at a particular set of
angles $\theta_{m}$ (where $m=0,1,2, \ldots )$ for red light of frequency
$f=4.6 \times 10^{14} \mathrm{Hz}$ . If one wishes to construct an analogous
two-slit interference set-up that produces interference fringes
at the same set of angles $\theta_{m}$ for room-temperature sound of
midde-C frequency $f_{\mathrm{s}}=262 \mathrm{Hz}$ , what should the slit sepa-
ration $d_{\mathrm{S}}$ be for this analogous set-up?

Key Concepts

Double-Slit Interference

Double-slit interference is a phenomenon observed when waves such as light or sound pass through two narrow, closely spaced openings. The waves emerging from the two slits overlap and interfere with each other, resulting in a pattern of alternating bright and dark fringes (or loud and soft regions in sound). This concept is fundamental in understanding wave behavior and the resulting interference effects that rely on the superposition principle.

Path Difference and Constructive Interference Condition

The fringe pattern in a double-slit experiment arises due to the difference in the paths traveled by the waves from each slit. Constructive interference, which produces the bright fringes (or louder sounds), occurs when the path difference is an integer multiple of the wavelength. This condition is mathematically expressed as d*sin(?) = m*?, where d is the slit separation, ? is the angle relative to the original direction, m is an integer (order of the fringe), and ? is the wavelength of the wave.

Wave Speed and Wavelength Relationship

A wave’s wavelength is directly related to its frequency and the speed of propagation through the medium, as given by the relation ? = v/f. For different types of waves or different media, the speed v can vary significantly, even if the frequency is known. This relationship is crucial when comparing interference patterns produced by different types of waves, as the wavelength determines the spacing and position of the interference fringes.

Scaling Experimental Parameters

When designing analogous interference experiments for different types of waves, adjusting experimental parameters such as the slit separation is necessary to maintain the same interference pattern. By understanding the relationship between the wavelength and the setup geometry (specifically, the slit separation that satisfies d*sin(?) = m*?), one can scale the slit separation according to the new wavelength. This ensures that the set of angles at which constructive interference occurs remains the same despite the difference in wave speeds and frequencies.

Transcript

00:01 Okay, so this is kind of a fun problem.

00:01 It gives you a situation for light interference, two -slet interference, and they want you to construct the same thing, but for a sound wave.

00:10 So i'm kind of interested in seeing what the result is.

00:13 So we know, let's start getting, let's get the givens down first.

00:17 We know that the frequency of the light is 4 .6 times 10 to the 14 hertz.

00:33 And we know the slit separates.

00:35 Is 0 .1 millimeters .1 times 10 to the minus 3 meters and okay and then we have a certain set of angles and we want that to be the case for f is 262 hearts so f s 22 so let's look at the equation that sets the the set of angles so we know that we know for two -slid interference that d times the sign of theta is equal to m -lamda.

01:26 So let's see.

01:33 So we know that the angle set is the same.

01:35 So then we can say sine theta.

01:44 Let's just bring all this stuff that has to do with this set of angles onto one side.

01:48 So we can say that that's equal to lambda over d.

01:53 So lambda over d is going to really set where our angles are.

02:01 Let me just stop and think a bit.

02:04 Okay, i think me thinking about it, i don't think it would make sense to move this m over.

02:09 What if we try to solve a simpler problem and just say that, look, we want their first angle to be the same, so that's m equals one.

02:15 And then i think, you know, if your first angle is the same, there's no reason why in this equation that your second and third angle wouldn't be the same because it just sort of scales.

02:24 With m.

02:26 And so let's just get the first angle to be the same.

02:31 Well, actually, i mean, we can just put this m back over here.

02:34 Just keep in mind if we want all the angles to be the same for the different m as 1 -2 -4, we need lambda over d to be the same.

02:41 So now let's figure out what lambda is in terms of frequency.

02:46 So some just kind of a side equation, frequency is c over lambda.

02:52 And c is the speed of the wave, lambda is the wavelength...