Diffraction grating spectrometer

In summary, to resolve the 589.00 and 589.59 nanometer lines in the second order (m=2), a reflection diffraction grating with n=140 lines per millimeter needs to be at least 3.565 millimeters wide.
  • #1
grouper
52
0

Homework Statement



Suppose that you have a reflection diffraction grating with n=140 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen. How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59 nanometers, which are a well known pair of lines for sodium, in the second order (m=2)?

Homework Equations



maxima for reflection diffraction grating: sinθ=(mλ)/d where m=2 in this instance

Rayleigh's resolution criterion: sinθ=1.22*(λ/d)

Spectrometer: λ=(d/m)*sinθ

d=distance between slits

The Attempt at a Solution



m=2 and d is the unknown. I tried setting the above equations equal to each other, eliminating sinθ, but I couldn't get any further because I don't know which λ to use since we are trying to separate two different wavelengths.
 
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  • #2
Use one λ, and then the other and compare.

Since they're so close together, the answers will be very nearly the same.
 
  • #3
Where am I supposed to use both λ's? Also, the question asks to create a situation in which the two λ's are far enough apart that they are indistinguishable, so they must be related somehow.
 
  • #4
589.00 and 589.59 nanometers
 
  • #5
Sorry, I meant I don't understand which equations to use and where to plug the two λ in.
 
  • #6
grouper said:

Homework Statement



Suppose that you have a reflection diffraction grating with n=140 lines per millimeter. Light from a sodium lamp passes through the grating and is diffracted onto a distant screen. How wide does this grating need to be to allow you to resolve the two lines 589.00 and 589.59 nanometers, which are a well known pair of lines for sodium, in the second order (m=2)?

Homework Equations



maxima for reflection diffraction grating: sinθ=(mλ)/d where m=2 in this instance

Rayleigh's resolution criterion: sinθ=1.22*(λ/d)

Spectrometer: λ=(d/m)*sinθ

d=distance between slits

The Attempt at a Solution



m=2 and d is the unknown. I tried setting the above equations equal to each other, eliminating sinθ, but I couldn't get any further because I don't know which λ to use since we are trying to separate two different wavelengths.
How is this different than the newer thread you started, except for the little extra you wrote:
"m=2 and d is the unknown. I tried setting the above equations equal to each other, eliminating sinθ, but I couldn't get any further because I don't know which λ to use since we are trying to separate two different wavelengths. I'm also a bit confused on which equation(s) is relevant. Also, I think my setting-them-equal strategy is flawed because d will always cancel out when you do that. I don't really know where to start."​
?

https://www.physicsforums.com/showthread.php?t=547558"
 
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  • #7
You need a relationship involving the total number of lines in the grating and the resolving power.
 
Last edited:
  • #8
Sorry, I was just trying to clarify what I have tried already. That post has been deleted. Our book doesn't have anything about resolving power and we haven't covered that in class (not sure why it's on a homework problem then). Someone in my class mentioned the equation sinθ=1.22(λ/d) but I'm not really sure what it applies to and it doesn't involve the number of lines. We covered diffraction grating in class but we haven't covered resolution at all, so I'm really lost on this problem. I'll look around online and see what I can learn.
 
  • #9
grouper said:
Someone in my class mentioned the equation sinθ=1.22(λ/d) but I'm not really sure what it applies to and it doesn't involve the number of lines.
That applies to a circular aperture; irrelevant here.

It's odd that your book and class lectures have not talked about this, yet you were assigned this problem. In any event, you might try a google search on: grating resolution
 
  • #10
Ok, I had to go outside the book to find the resolution equations R=λ/Δλ and for a spectrometer R=mN where N=d*n and setting R=R got d=3.565 mm, which was correct.

It is pretty frustrating that our book contains a question about resolution and yet doesn't even mention those equations. I think I might write to the publisher about that. Thanks for the help everyone! At least now I understand resolution for the next test!
 

What is a diffraction grating spectrometer?

A diffraction grating spectrometer is a scientific instrument used to measure the spectrum of light. It consists of a diffraction grating, which is a finely etched surface that separates light into its component wavelengths, and a detector, which measures the intensity of each wavelength of light.

How does a diffraction grating spectrometer work?

A diffraction grating spectrometer works by directing a beam of white light onto a diffraction grating. As the light passes through the grating, it is diffracted into its component wavelengths. The diffracted light is then focused onto a detector, which measures the intensity of each wavelength. This data is then used to create a spectrum of the light source.

What are the advantages of using a diffraction grating spectrometer?

There are several advantages to using a diffraction grating spectrometer. Firstly, it has a high resolution, meaning it can distinguish between very closely spaced wavelengths of light. It is also a relatively inexpensive and compact instrument compared to other spectrometers. Additionally, it can be used to measure a wide range of wavelengths, from ultraviolet to infrared.

What are some common applications of a diffraction grating spectrometer?

A diffraction grating spectrometer has many applications in various fields of science and engineering. It is commonly used in astronomy to study the composition of stars and other celestial objects. It is also used in chemistry to identify and analyze the components of a sample. In physics, it can be used to study the properties of light and how it interacts with matter.

How do you calibrate a diffraction grating spectrometer?

The calibration of a diffraction grating spectrometer involves using a known light source, such as a laser, to determine the relationship between the wavelength of light and the position of the diffracted light on the detector. This calibration can then be used to accurately measure the spectrum of an unknown light source.

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