Double Slit Interference: Wavelength & Central Maximum

[Soaring Crane]

Two lasers are shining on a double slit, with slit separation d. Laser one has a wavelength of d/20, while laser two has a wavelength of d/15 . The lasers produce separate interference patterns on a screen a large distance L away from the slits.

Which laser has its first maximum closer to the central maximum?

I read that the first maximum comes when the path difference between the two slits is equal to one full wavelength. What does this have to do with the laser that has a smaller wavelength (laser 1)? The distance from the central maximum is proportional to the path-difference, but I stiill don't understand exactly what the central maximum is.

Any help is appreciated.

 

[The Bob]

Two lasers are shining on a double slit, with slit separation d. Laser one has a wavelength of d/20, while laser two has a wavelength of d/15 . The lasers produce separate interference patterns on a screen a large distance L away from the slits.

Which laser has its first maximum closer to the central maximum?

I read that the first maximum comes when the path difference between the two slits is equal to one full wavelength. What does this have to do with the laser that has a smaller wavelength (laser 1)? The distance from the central maximum is proportional to the path-difference, but I stiill don't understand exactly what the central maximum is.

Any help is appreciated.

Laser one will have the first maximum that is closest to the bright centre.

The central maximum is the point where the light shins straight through the slits and hits the screen. This is when the path difference is 0.

As the path difference of the next bright point is when the wavelengths are one out of each other the shorter laser will be constructively interferred with first. The longer wavelength will do the same but further away.

If you are really stuck, apply numbers into the problem and see what you find or ask some more questions. I don't know what it is you are not understanding.

The Bob (2004 ©)

 

[Soaring Crane]

How do you find the minimum of a laser in terms of L? I am trying to find the third minimum's location of laser 2 and I know m is 2, but I don't know what formula to use aside from y = [m(lambda)*L]/d, which is for the maximum or 2L/15. What do I do?

 

[jtbell]

For two-slit interference, the minima are located halfway between the maxima. You can locate them by using half-integers for m: 0.5, 1.5, 2.5 etc.

 

[Soaring Crane]

I'm not sure that I understand this method. I better type out the whole problem, so you can see my current solutions.

What is the distance between the second maximum of laser one and the third minimum of laser two, on the same side of the central maximum?

If the central maximum corresponds to m = 0 , then you should be able to figure out what the second maximum corresponds to. Using that, the distance to the second maximum of laser one from the central maximum is L/10 or L/9.95.

The first minimum corresponds to m = 0 (since there is no central minimum). The value of m for the third minimum is m = 2 (m is always an integer).

Now the part that I am trying to figure out:
Given that m = 2, what is the location of the third minimum?

Once I find this value, I can subtract this from the second maximum to find the distance between both locations.

 

[jtbell]

The first minimum corresponds to m = 0 (since there is no central minimum). The value of m for the third minimum is m = 2 (m is always an integer).

That means they're assuming the following formula for the minima:

y = \frac {(m + 1/2) \lambda L} {d}

 

[Soaring Crane]

So I must use this to find the third minimum's location?

 

[jtbell]

In the formula I gave you, m = 0 gives you the first minimum (on either side of the central maximum), m = 1 gives you the second minimum, etc.

Notice the pattern:

0(\lambda L / d) gives the central (zero'th) maximum.

0.5 (\lambda L / d) gives the first minimum away from the center.

1 (\lambda L / d) gives the first maximum away from the center.

1.5 (\lambda L / d) gives the second minimum away from the center.

2 (\lambda L / d) gives the second maximum away from the center.

Etc.