Seperation for third order maxima for two wavelengths - help please

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Homework Help Overview

The problem involves a diffraction grating with 1070 lines per centimeter and a screen positioned 3.20 m away. The original poster is attempting to calculate the separation between the third-order maxima for two wavelengths of light, 630 nm and 705 nm, using the diffraction equation.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to calculate the grating spacing and the angles for the third-order maxima using the diffraction equation. There are questions regarding the arithmetic involved in determining the grating spacing and the correct application of the formula. Some participants suggest checking calculations and units, while others question the assumptions made in the setup.

Discussion Status

The discussion is ongoing, with participants providing feedback on arithmetic errors and unit conversions. There is a focus on ensuring the calculations are correct and that the original poster is using the appropriate significant figures. Multiple interpretations of the problem are being explored, particularly regarding the calculations and the expected units for the final answer.

Contextual Notes

Participants note potential issues with significant figures and the specific requirements of the online submission system. There is also mention of the need for clarity on the problem statement to avoid ambiguity in the calculations.

tg22542
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Homework Statement




A grating has 1070 lines per centimetre, and a flat screen is perpendicular to the ray that makes the central peak of the diffraction pattern. The screen is 3.20 m from the grating. If light of two wavelengths, 630 nm and 705 nm, passes through the grating, what is the separation on the screen between the third-order maxima for the two wavelengths?


Homework Equations



mλ = dsinΘ

The Attempt at a Solution




d = 1/1070 = .000935 * 10 000 000 = 9345.79nm

λ1 = Θ1 = sin^-1(3*705 / 9345.79) = 13.0796°
λ2 = Θ2 = sin^-1(3*630 / 9345.79) = 11.6674°

[3.2tan(13.0796°)] - [3.2tan(11.6674)] = .082673 * 1000 = 82.67 nm


My answer is incorrect, can anyone see where I'm going wrong here?

Also, where I multiplied my numerator by 3 above, I did that due to the third order maxima, is that correct? I was impatient and also tried multiplying it by 2 instead, also incorrect.

Please help me out here.
 
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tg22542 said:
A grating has 1070 lines per centimetre,
Check your arithmetic.
 
oops.

Then shouldn't using (1/107) * 10000000 as my new value be right?

It's still wrong.
 
tg22542 said:
Then shouldn't using (1/107) * 10000000 as my new value be right?

It's still wrong.
Two things:
(1) That should be (.1/1070)
(2) 1 nm = 10^-9 m
 
I still got it wrong.

I'll show you once more what I'm doing nowd = .1/1070 = .000093 * 10 000 000 = 934.579nm

λ1 = Θ1 = sin^-1(3*705 / 934.579) = x1°
λ2 = Θ2 = sin^-1(3*630 / 934.579) = x2°

[3.2tan(x1°)] - [3.2tan(x2)] = z

z*1000 = ansstill wrong thoughWhere exactly do i need to make changes, and what are those changes? I have 6 tries left.
 
tg22542 said:
I still got it wrong.

I'll show you once more what I'm doing nowd = .1/1070 = .000093 * 10 000 000 = 934.579nm
.1/1070 = 0.0000934579 m. So far, so good.
Why do you then multiply by 10 000 000? Instead, multiply by 109.
 
im doing all of this and still getting it wrong
 
tg22542 said:
im doing all of this and still getting it wrong
Post each step, just like you did before.
 
Most previous try:

d = .1/1070 = .000093 * 10^9 = 93457.9 nm

λ1 = Θ1 = sin^-1(3*705 / 93457.9) = 1.297°
λ2 = Θ2 = sin^-1(3*630 / 93457.9) = 1.159°

[3.2tan(1.297)] - [3.2tan(1.159)]

= .0077 nm
 
  • #10
tg22542 said:
Most previous try:

d = .1/1070 = .000093 * 10^9 = 93457.9 nm

λ1 = Θ1 = sin^-1(3*705 / 93457.9) = 1.297°
λ2 = Θ2 = sin^-1(3*630 / 93457.9) = 1.159°

[3.2tan(1.297)] - [3.2tan(1.159)]

= .0077 nm
That answer in is meters, not nm. What units do they want the answer in? mm makes sense.
 
  • #11
Well the units don't matter, hold on I will convert it and give it a try!
 
  • #12
7.7mm didn't work
 
  • #13
tg22542 said:
7.7mm didn't work
I don't see an obvious error, unless it's just a significant figures issues. Often those systems want 3 sig figs no matter what.

Otherwise I will look at this more carefully later and see if there's another error that I missed.
 
  • #14
Sounds great. Thank you very much for your help this far.
 
  • #15
tg22542 said:
Sounds great. Thank you very much for your help this far.
I don't see any errors in the calculation.

Did you try using 3 sig figs in your response? (I assume this is some online system, not a textbook problem, right?)

If all else fails, it might help to take a screenshot of the question so that we can check that there's no ambiguity. (Though it seems pretty clear.)
 
Last edited:
  • #16
I did try 3 sig figs.. still wrong. In my angle calculation (second step), I multiplied each wavelength by 3, is that correct?
 
  • #17
tg22542 said:
In my angle calculation (second step), I multiplied each wavelength by 3, is that correct?
Sure.
 

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