# Diffraction Grating, calculating rulings/mm from spectrum

• Oijl
In summary, the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of 20.0 degrees is not explicitly solvable, but a numerical approach can be used to find the value of d.
Oijl

## Homework Statement

Assume that the limits of the visible spectrum are arbitrarily chosen as 430 and 680 nm. Calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of 20.0 degrees.

## Homework Equations

(y is lambda)
dsinø = my (maxima)

## The Attempt at a Solution

So, the first-order spectrum of white light has a line (maxima) the lowest wavelength and another line at the highest wavelength - right?

And for the first-order spectrum would have a m of 1, right?

If this is true, then,
dsin(ø1) = 430
dsin(ø2) = 680

And, ø2 - ø1 = 20 degrees

And
ø1 = arcsin(430/d)
ø2 = arcsin(680/d)

But, I want to find out d. How can I extract d from these equations?

Or, alternatively, do I not understand what's going on?

You have the right approach, but we can't solve for d explicitly; a numerical approach is needed here. What value of d results in ø1 and ø2 that are 20 degrees apart?
If you can set this up in a spreadsheet, it won't take very long to find d that gives the required 20 degree spread.

Hello,

Your understanding of the first-order spectrum is correct, with the first maximum occurring at the lowest wavelength and the second maximum at the highest wavelength. However, you are correct in wanting to solve for d, the number of rulings per millimeter, in order to calculate the number of rulings per millimeter of a grating that will spread the first-order spectrum through an angle of 20.0 degrees.

To solve for d, you can use the equation dsinø = my (maxima) and plug in the values you have for ø1 and ø2. This will give you two equations with two unknowns (d and y). You can then solve for d by setting the two equations equal to each other and simplifying. This will give you the value of d in terms of the other variables.

Alternatively, you can also use the equation ø2 - ø1 = 20 degrees to solve for d. By rearranging the equation to solve for d, you can then plug in the values for ø1 and ø2 and solve for d. Both methods should give you the same value for d. I hope this helps!

## 1. What is a diffraction grating?

A diffraction grating is a device that consists of a large number of parallel and evenly spaced slits or lines. When light passes through the grating, it is diffracted into multiple beams, creating a spectrum of colors. This is due to the interference of light waves passing through the slits.

## 2. How is the number of rulings per millimeter calculated?

The number of rulings per millimeter (r/mm) on a diffraction grating can be calculated by dividing the number of lines or slits on the grating by the width of the grating in millimeters. This value represents the spacing between each line on the grating and is used to determine the diffraction angle of the light passing through.

## 3. What is the relationship between the diffraction angle and the number of rulings per millimeter?

The diffraction angle (θ) is directly proportional to the number of rulings per millimeter (r/mm) on a diffraction grating. This means that as the number of rulings per millimeter increases, the diffraction angle also increases, resulting in a broader spectrum of light.

## 4. How does the material and spacing of the lines on a diffraction grating affect its performance?

The material and spacing of the lines on a diffraction grating can affect its performance by altering the intensity and wavelength range of the diffracted light. Different materials and spacing can also result in varying levels of dispersion, which is the separation of different wavelengths of light in the spectrum.

## 5. What are the practical applications of diffraction gratings?

Diffraction gratings are commonly used in spectroscopy, which is the study of the interaction between matter and electromagnetic radiation. They are also used in the production of holograms, optical filters, and in laser systems for wavelength selection. They are also important in fields such as astronomy, chemistry, and medicine.

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