How Many Lines Per Millimeter Does the Diffraction Grating Have?

[Ericv_91]

Homework Statement

The figure shows the interference pattern on a screen 1.0 m behind a diffraction grating. The wavelength of the light is 600 nm. How many lines per millimeter does the grating have?

Homework Equations

lines/mm = 1/ distance between slits

The Attempt at a Solution

Quite frankly I don't have a clue how to solve this. I know that the number of lines per millimeter is equal to the inverse of the distance between slits, but what happens if there are more than 2 slits of different distances from each other? I took a shot in the dark and tried to add up the reciprocals of all the distances, but that proved to be incorrect because it gave me a very small number around 0.006. Does anyone have any helpful suggestions?

 

[vela]

Quite frankly I don't have a clue how to solve this. I know that the number of lines per millimeter is equal to the inverse of the distance between slits, but what happens if there are more than 2 slits of different distances from each other? I took a shot in the dark and tried to add up the reciprocals of all the distances, but that proved to be incorrect because it gave me a very small number around 0.006. Does anyone have any helpful suggestions?

It's hard to say where you went astray because you didn't show your work.

I'm guessing you used $y_1=0.436~\rm m$ and $y_2=0.897~\rm m$ and ended up getting to different answers for $d$. The problem is that 0.897 m is the distance between the first- and second-order maxima, so you need to use $y_2 ={0.897~\rm m}+y_1$.

 

[Steve4Physics]

I see @vela has already beaten me to it by 20 minutes or so. But given I’ve already drafted the following, I’ve decided to post it!

I think the OP (who will probably never read this thread) mistakenly thought that the uneven spacing of the maxima implied different line-spacings.
__________

This is a 12+ year old question, but in case the answer is of use to someone…

From the data in the Post #1 diagram:

The first order maximum makes an angle θ₁ with the normal, where $θ₁ = tan⁻¹ (\frac {43.6}{100}) = 23.56º$.

The diffraction grating formula (nλ = dsinθ) with n=1 and θ = θ₁ gives:
$d = \frac {1*600e-9}{sin(23.56º)}$
If we express d in units of mm, then ‘lines/mm’ = $\frac 1 d$.

To check this value, we can repeat the process using the second order maximum.

The second order maximum is a distance 89.7cm+43.6cm = 133.3cm from the axis. So $θ₂ = tan⁻¹ (\frac {133.3}{100}) = 53.12º$.

Using ‘nλ = dsinθ’ with n=2 and θ = θ₂ will give the same value for ‘lines/mm’ as previously calculated.

 

[Sanrasz]

To add to this, d is in m when you calculate it, so multiply the resulting d value by 1000 to convert to mm, then divide 1 by the result to get the answer. (I didn't see that part and was slightly confused)

 

[kuruman]

To add to this, d is in m when you calculate it, so multiply the resulting d value by 1000 to convert to mm, then divide 1 by the result to get the answer. (I didn't see that part and was slightly confused)

Thank you for your contribution. Please note that this thread is more than two years old. The OP is unlikely to profit from it at this point.

 

[Sanrasz]

Thank you for your contribution. Please note that this thread is more than two years old. The OP is unlikely to profit from it at this point.

I don't particularly care about the OP, but this question is actually extremely common in homework problems of the fifth edition of the book that OP's question is from. (And is exactly why I was reading this thread. Everything, down to the figure matches.)