# Finding the Distance between fringes given different wavelengths

1. Feb 19, 2013

### rocapp

1. The problem statement, all variables and given/known data
See attached image

2. Relevant equations
single wavelength
mλ=dsin(θ)

diffraction grating:
d*(sinθi+sinθm)=mλ

y/L = sinθ

3. The attempt at a solution

red = 656 nm
blue = 486 nm
L = 1.7 m

I really do not know what to do.

I tried 656-486 = 170 nm

so
y/L = sinθ
0.000017 cm/ 170 cm = sinθ

θ=5.73x10^-6

and that does not yield a correct answer.

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Last edited: Feb 20, 2013
2. Feb 20, 2013

### Simon Bridge

The attachment didn't make it.
Work out the spacing for the fringes of each wavelength separately.

3. Feb 20, 2013

### rocapp

Which equation should I use? I realize that I can plug in lambda for each wavelength into two different equations, but I don't see how I can use the given information to find d. I have no y-values. I have never used the diffraction-grating formula, but it requires an incident and reflecting value for theta and I do not have these.

4. Feb 20, 2013

### Simon Bridge

All right - the situation is that you have a diffraction grating with 550lines/mm, a screen 1.7m from the grating, and the red and blue hydrogen discharge lines.

You want to find the distance between the 1st order fringes.
In order to use equations in physics you have to understand them. You should go back over your notes about how the fringes form in the first place.

In the equation $d\sin\theta = n\lambda$ - the $d$ is the distance between the lines on the diffraction grating, $\lambda$ is the wavelength of the light, $n$ is the "order" of the fringe, and $\theta$ is the angle from the center of the grating to the fringe.

You can find the angle to each fringe with this formula - but you need some way to turn those angles into distances on the screen.

5. Feb 20, 2013

### rocapp

0.00002*sin(θ) = 656x10-6mm
sin(θ) = 32.8
sin^-1(32.8) = θ

θ = 90°

0.00002*sin(θ) = 486x10-6mm
sin(θ) = 24.3
θ = 90°

Every time I have gotten a right angle with this formula so far, I have been wrong... haha.

My next step would be:

y = L*tanθ
y = (1700 mm)*tan(90) = ... obviously I have done something wrong.

I tried again using y = (lambda*L)/d

and found 15879.4 cm distance, but that was incorrect.

Last edited: Feb 20, 2013
6. Feb 20, 2013

### Simon Bridge

so you think that d=0.00002mm huh?
If you have 550 lines per millimeter - how many millimeters do you have per line?
That cannot be right: the sine must always be less than 1!
So the next bit
... is saying is that sin(90)=32.8 : is that correct?

7. Feb 20, 2013

### rocapp

For the first step, I would say that since 1/500=0.00002, that would be correct.

EDIT
my mistake! Geez I must have been tired when I was doing that. The correct answer is:

1/550=0.00182 mm

So in cm:

(1.82x10^-7)*sin(theta) = (6.56x10^-5)
sin(theta) = 360.44

theta = 90 degrees.

Still getting it. :P

Last edited: Feb 20, 2013
8. Feb 21, 2013

### Simon Bridge

You are still making the same mistakes: focus!
Your numbers are describing a right-angled triangle whose hypotenuse is shorter than the opposite side.
It cannot happen... geometry won't allow it.

(1.82x10^-7)*sin(theta) = (6.56x10^-5)
... Include your units in the calculation and redo.

Last edited: Feb 21, 2013
9. Feb 26, 2013

### rocapp

Thank you! I have had a terrible time just calming down enough to pay attention to the specifics of what I am doing.

So dsin(theta) = m(lambda)

1.82x10^-3 mm * sin(theta) = 1 * 6.56x10^-4 mm
sin (theta) = 0.360
theta = arcsin(0.360)
theta = 21.13 degrees

Last edited: Feb 26, 2013
10. Feb 26, 2013

### rocapp

for the blue maximum's distance

d*sin(theta) = m*(lambda)
(1.82x10^-3 mm)*sin(theta) = 1*(4.86x10^-4 mm)
arcsin()

For the red maximum's distance

y = L*tan(theta)

y(red) = 1.7*tan(21.13)

y(red) = 0.657 m

theta blue = 15.49 degrees

y blue = 0.471 m

delta y = d = 0.186 m

So the distance between the two maxima should be 0.186 m.

Please let me know if this answer is correct; it is due tomorrow.

Last edited: Feb 26, 2013
11. Feb 28, 2013

### Simon Bridge

That is much better - see how this kind of discipline works out?
To reality-check your own answers - do what I was doing before: work out what they mean.

Before: your data was telling you that the hypotenuse of a right-triangle was smaller than the other sides - which does not make sense, therefore you were probably wrong. After a while you get used to thinking this way: the problem is not over just because you have finished the calculation.

This last result means that if you fed white light into the diffraction grating, you get a white central maxima, and a series of rainbow-colored fringes to either side. The red part will be farther away from the center than the blue part, and the individual fringes will be about 186mm wide. If all this makes sense then you have probably go it right.

Just an aside - normally, where L >> d (as in this case) then $\tan\theta \approx \sin\theta \approx \theta$ (using radians instead of degrees) and you can say that $n\lambda/d \approx x/L$ and you can do it in one step.

Since $x=\frac{L}{d}\lambda$ then $\Delta x = \frac{L}{d}(\lambda_{red}-\lambda_{blue})$

Do get used to using radians by default for angles.

Last edited: Feb 28, 2013
12. Mar 4, 2013

Thanks!