# Measuring the Distance between the Fringes of a Diffraction Grating

KDPhysics
Homework Statement:
Determining the pitch distance on a CD acting as a transmission grating.
Relevant Equations:
$$d\sin\theta = n\lambda$$
I have been preparing for a physics practical on diffraction. More specifically, we will use a CD as a transmission grating (by peeling off the reflective layer), and measure the distance between the fringes for a specific distance between the CD and the viewing wall. However, it is unclear whether we should measure the distance between the two 1st order fringes and divide by two, or measure the distance from the principal maxima to the left 1st order, and then from the principal maxima to the right 1st order, and take the average. Which is more suitable for this experiment? I believe that there is no great difference, since the absolute uncertainty remains the same in both cases.

Homework Helper
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What do you mean by absolute uncertainty? If you want to find the radius of a coin do you find the center first and then measure from the center of the coin to the edge or do you find the distance along a diameter and divide by two?

• KDPhysics
KDPhysics
Well, let the measurements for the 1st order right fringe be ##x_r\pm\delta x## and the measurement to the left be ##x_l\pm\delta x##. Taking the average of the two: ##\hat{x} = \frac{x_r+x_l}{2}\pm\delta x##.
If instead we measure the distance from one fringe to the other, and then divide by two: ##x = \frac{x_r+x_l}{2}\pm\delta x## which is the same.

Homework Helper
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Well, let the measurements for the 1st order right fringe be ##x_r\pm\delta x## and the measurement to the left be ##x_l\pm\delta x##. Taking the average of the two: ##\hat{x} = \frac{x_r+x_l}{2}\pm\delta x##.
If instead we measure the distance from one fringe to the other, and then divide by two: ##x = \frac{x_r+x_l}{2}\pm\delta x## which is the same.

• KDPhysics
KDPhysics
yeah should be fixed sorry

KDPhysics
What do you mean by absolute uncertainty? If you want to find the radius of a coin do you find the center first and then measure from the center of the coin to the edge or do you find the distance along a diameter and divide by two?
By absolute uncertainty, I mean the sensitivity in the measurement (if I were using a ruler, it would be ##\pm 0.001m##).
Also, in the case of the coin you usually don't have the exact center. Instead, in the diffraction pattern you do, it's the principal maxima.

Homework Helper
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By absolute uncertainty, I mean the sensitivity in the measurement (if I were using a ruler, it would be ±0.001m±0.001m).
Thanks for fixing the LaTeX. Uncertainties don't add the way you show. You need to consider error propagation. See here for example. In the second procedure you add an extra uncertainty due to the measurement of the central maximum.

On edit: I used the coin example to illustrate that the determination of the radius is more accurate when you eliminate the uncertainty of finding the position of the center which is what you have to do here.

• KDPhysics
KDPhysics
Shouldn't the uncertainty be the same either way? Measuring from one fringe to the other, then the uncertainty will be 0.002m using a meter ruler. Similarly, measuring from the principal maxima to each fringe, the uncertainty will still be 0.002m.

Also, we can use the linear approximation for error propagation (since we're still in high school).

EDIT: measurement uncertainty should be ##\pm0.001m##

Last edited:
KDPhysics
Both in the case of measuring from one fringe to the other, or from the principal maxima to each fringe, we're still measuring the distance between two points, so shouldn't the uncertainty be the same for both measurements?

Homework Helper
Gold Member
Shouldn't the uncertainty be the same either way? Measuring from one fringe to the other, then the uncertainty will be 0.002m using a meter ruler. Similarly, measuring from the principal maxima to each fringe, the uncertainty will still be 0.002m.

Also, we can use the linear approximation for error propagation (since we're still in high school).

• KDPhysics
KDPhysics
We're allowed to use the second formula (##\delta R = \delta X + \delta Y + \delta Z##).

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Homework Helper
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We're allowed to use the second formula (##\delta R = \delta X + \delta Y + \delta Z##).
OK then, but let me ask you this. Suppose that you measure the two first order maxima from the center and you find that one is at a greater distance than the other by more than your assumed uncertainty. What will you conclude from that?

• KDPhysics
KDPhysics
Ok, I think I got it. Since one is slightly farther away from the other, we should calculate the half range:
$$\frac{x_r - x_l}{2}>\delta x$$
assuming the right measurement is greater than the other. Then, we should be using this value as our uncertainty, which is greater than ##\delta x##. We can therefore conclude that measuring from one fringe to the other will decrease uncertainty.

Is this correct?

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