Solving Signal Intensity Homework: Lines per mm & Diffraction Angle

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Homework Help Overview

The discussion revolves around designing a diffraction grating to disperse the visible spectrum over a specified angle. The problem includes determining the number of lines per millimeter needed for the grating and calculating the diffraction angle for a specific wavelength of light.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the implications of using two different wavelengths and how to approach the calculation of the grating's line density. There is uncertainty regarding whether to calculate the grating spacing for both wavelengths or to find a single value that relates to the overall spread.

Discussion Status

Some participants have suggested looking at the problem from a derivative perspective to understand the relationship between wavelength and angle. Others are questioning which value of grating spacing (d) should be used for the calculations, indicating a lack of consensus on the approach to take.

Contextual Notes

Participants note the challenge of interpreting the problem correctly, especially with the presence of multiple wavelengths and the specified angle. There is mention of the need to consider how changes in wavelength affect the diffraction angle, which adds complexity to the calculations.

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Homework Statement



You need to design a diffraction grating that will disperse the visible spectrum (400-700nm) over 30.0 degree in first order.
a) How many lines per millimeter does your grating need?
b) What is the first-order diffraction angle of light from a sodium lamp (wavelength = 589nm)?

Homework Equations



dsin(theta) = m(wavelength)

The Attempt at a Solution



m= 1 (as it is first order) wavelength = (400*10^-9)m
angle = 30 degree wavelength = (700*10^-9) m


I'm confused because there are two wavelengths given. Do I find the d value for both wavelength and then find difference between them ... d= 0.0000014-0.0000008 = 0.0000006 = 6*10^-7
then 1667lines/mmm

for part 2) do i use the d value as 6*10^-7
because other values are given ... wavelength = 589 *10^-9 and m=1


If I take the same d value I'm getting as 79 degree ... is it right ?

Please somebody check the calculation and tell if its right ... I've only one attempt left.

Can someone also tell me how to put equation on these kind of questions. I'll be very grateful.
 
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This may help

Howdy, I'm not 100% sure, but I think what you need to do is look at it from the point of view of rates of change.

So, from dsin(theta) = m(wavelength), you first need to find the formula for the derivative d(wavelength)/d(theta). Then set it equal to (change in wave/change in theta) from the given info and solve for d in mm. Then take the reciprical like you did to see how many d in 1 mm. For part 2 you just plug in 589, after taking account of units of course, and solve for theta.

AI hope that helps.
 
AtomicBartender said:
Howdy, I'm not 100% sure, but I think what you need to do is look at it from the point of view of rates of change.

So, from dsin(theta) = m(wavelength), you first need to find the formula for the derivative d(wavelength)/d(theta). Then set it equal to (change in wave/change in theta) from the given info and solve for d in mm. Then take the reciprical like you did to see how many d in 1 mm. For part 2 you just plug in 589, after taking account of units of course, and solve for theta.

AI hope that helps.

which value of d should take in part 2
 
pari786 said:
which value of d should take in part 2

After completing step 1, and imputing the initial data, there is only 1 d for the system. So, use the d "from the given info and solve for d in mm" (from my 1st post). Strategically, looking at the written problem and supporting equation, and seeing the 400nm and 700 nm givens, makes some people just want to rush ahead and plug them in and solve for each d and go from there, rather, in this case, we are interested in how a change in wavelength (their difference) is related to a change in spread (theta).
 
The derivative works for small changes, but to be more accurate you'll have to solve the full equations. Neither wavelength1 nor wavelength2 have theta = 30 degrees, the question asks what value of d gives a difference in theta values equal to 30 degrees. You know d*sin(theta1) = wavelength1, and you know d*sin(theta1 + 30 degrees) = wavelength2, so you can solve for theta1 and d in part a. Given d, you can then find the value of theta corresponding to 589 nm in part b.
 

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