Need help with diffraction problem by tonight

[eku_girl83]

Here's the problem:
Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two wources are coherent, 1.72e-6 m apart, and in line with an observer, so that one source is 1.72e-6m farther from the observer than the other. What is the longest of the visible wavelengths (400 to 700 nm) at which the observer will see the brightest light, owing to constructive interference?

I read the section in the book that problem came from, but the only equation I can find is r2-r1=m*lambda (constructive interference, sources in phase). How do I apply this equation to this problem?

This is probably my last physics question, since our final is Monday. I just want to thank everyone on this board. You have been so helpful to me...I don't know how I would have gotten my homework done without your help!

 

[gnome]

Better if you don't think in terms of "using an equation".

Constructive interference occurs where the difference in path length (the difference between the distances that light from the two sources travels) is an integer multiple of the wavelength of the light: 1 x the wavelength, or 2 x the wavelength, or 3 x ... etc.

In your example, that difference is given as 1.72 x 10^(-6) m which is equal to 1720 nm. Looking at your choices of wavelengths, you can see that 1720 isn't 1 times any of them, or 2 times any of them. But it is 3 times one of them: 1720/3≅573, so at that distance there will be constructive interference for light of 573nm wavelength.

If you insist on using the equation: r2 - r1 is the path-length difference: 1720 nm.
m is 3. λ is the (unknown) wavelength. You start plugging values into m, starting with m=1, then try m=2 etc, until you find an "m" that works. Note that m=4 works too, for light of wavelength 430 nm. But it won't be as bright as the constructive interference at m=3.

 

[M98Ranger]

Hi there,

Thank you for reaching out for help with your diffraction problem. I am happy to assist you with understanding and solving this problem.

Firstly, let's break down the information given in the problem. We have two light sources that can emit monochromatic light (light of a single wavelength) of any visible wavelength (400 to 700 nm). These sources are coherent, which means they have a constant phase relationship. They are also 1.72e-6 m apart and in line with an observer, with one source being 1.72e-6 m farther from the observer than the other.

The question is asking for the longest visible wavelength at which the observer will see the brightest light due to constructive interference. This means that we need to find the wavelength at which the two light sources will add up in phase to create the brightest light.

Now, let's look at the equation you mentioned, r2-r1=m*lambda. This is the equation for constructive interference, where r2 and r1 are the distances from the two light sources to the observer, m is the order of the interference (1, 2, 3, etc.), and lambda is the wavelength of the light.

In this problem, we know that r2-r1 is 1.72e-6 m, which is the distance between the two light sources. We also know that m=1, since we are looking for the first order of interference (the brightest light). Therefore, we can rearrange the equation to solve for lambda, which gives us lambda = (r2-r1)/m.

Plugging in the values we know, we get lambda = (1.72e-6 m)/1 = 1.72e-6 m. This means that the longest visible wavelength at which the observer will see the brightest light is 1.72e-6 m, which is equivalent to 1720 nm. This is outside of the visible spectrum, which ranges from 400 to 700 nm. Therefore, the observer will not see the brightest light at any visible wavelength.

I hope this explanation helps you understand and solve the problem. If you have any further questions, please feel free to ask. And best of luck on your final exam! You got this.