Solving Signal Intensity Homework: Lines per mm & Diffraction Angle

  • Thread starter Thread starter khushi08
  • Start date Start date
  • Tags Tags
    Intensity Signal
Click For Summary
SUMMARY

The discussion focuses on designing a diffraction grating to disperse the visible spectrum (400-700 nm) over a 30-degree angle in first order. The required number of lines per millimeter is calculated to be 1667 lines/mm based on the derived value of d, which is determined from the difference in wavelengths. For the second part, the first-order diffraction angle for a sodium lamp with a wavelength of 589 nm is calculated using the same d value, leading to a result of approximately 79 degrees. The discussion emphasizes the importance of correctly applying the diffraction equation dsin(theta) = m(wavelength) and understanding the relationship between wavelength changes and angular spread.

PREREQUISITES
  • Understanding of diffraction grating principles
  • Familiarity with the equation dsin(theta) = m(wavelength)
  • Basic knowledge of calculus, specifically derivatives
  • Ability to manipulate units in physics calculations
NEXT STEPS
  • Study the derivation of the diffraction grating equation dsin(theta) = m(wavelength)
  • Learn how to calculate diffraction angles for multiple wavelengths
  • Explore the concept of derivatives in the context of wave phenomena
  • Investigate practical applications of diffraction gratings in spectroscopy
USEFUL FOR

Students in physics or engineering courses, optical engineers, and anyone involved in designing or analyzing diffraction gratings for spectral analysis.

khushi08
Messages
1
Reaction score
0

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.
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
992
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K