Diffraction Grating wavelength

[ReidMerrill]

Homework Statement

A) Find the diffraction order from the grating that has an angle of incidence of 40 degrees and a diffraction angle of 20 degrees. The wavelength of incident light is 500nm and the spacing between the grooves on the grating are 5 micrometers
B) If I want to selectively emit 500 nm wavelength from the grating, what is the actual range of wavelengths that would be diffracted if the grating had 6500 grooves.

I was able to answer part A easily but I can figure out part B. and that's no use.

What is/are the relevant formula(s)?

Homework Equations

This is for a chemistry class and I can't find any relevant formula. The only ones from the book are snells law and n(Lambda)=d(Sin(theta)+sin(Phi))

The Attempt at a Solution

 

[Charles Link]

The resolving power $R$ of a grating is $R=\frac{\lambda}{\Delta \lambda}=Nm$ where $N$ is the number of lines(grooves) on the grating and $m$ is the order of the maximum (the "n" in your formula above) . ($m=1$). You need to solve for the resolution $\Delta \lambda$ which is the (approximate) spread of wavelengths. The range is then given approximately by $\lambda-\frac{\Delta \lambda}{2} < \lambda< \lambda+\frac{\Delta \lambda}{2}$ $\\$ (Additional note: this is the maximum resolving power and minimum resolution that the spectrometer can achieve. If you use wide slit widths on your spectrometer, you will find the resolving power decreases and the spread of wavelengths emerging from the exit slit increases. This extra detail is beyond the scope of what your textbook addresses with this homework question, but you may find it of interest. The relevant formula in this case is $\Delta \lambda=d (\Delta x)/f$ where $\Delta x$ is the slit width and $f$ is the focal length of the spectrometer's focusing optics. In order to achieve maximum resolving power, the slit width needs to be sufficiently narrow that this second $\Delta \lambda$ is smaller than the $\Delta \lambda$ that you computed above. Most often it is the slit width that determines the observed spread $\Delta \lambda$, because the experimentalist needs to use slit widths that are wide enough to get sufficient amount of light at the exit slit for easy observation.)

 

[Charles Link]

The resolving power $R$ of a grating is $R=\frac{\lambda}{\Delta \lambda}=Nm$ where $N$ is the number of lines(grooves) on the grating and $m$ is the order of the maximum (the "n" in your formula above) . ($m=1$). You need to solve for the resolution $\Delta \lambda$ which is the (approximate) spread of wavelengths. The range is then given approximately by $\lambda-\frac{\Delta \lambda}{2} < \lambda< \lambda+\frac{\Delta \lambda}{2}$ $\\$ (Additional note: this is the maximum resolving power and minimum resolution that the spectrometer can achieve. If you use wide slit widths on your spectrometer, you will find the resolving power decreases and the spread of wavelengths emerging from the exit slit increases. This extra detail is beyond the scope of what your textbook addresses with this homework question, but you may find it of interest. The relevant formula in this case is $\Delta \lambda=d (\Delta x)/f$ where $\Delta x$ is the slit width and $f$ is the focal length of the spectrometer's focusing optics. In order to achieve maximum resolving power, the slit width needs to be sufficiently narrow that this second $\Delta \lambda$ is smaller than the $\Delta \lambda$ that you computed above. Most often it is the slit width that determines the observed spread $\Delta \lambda$, because the experimentalist needs to use slit widths that are wide enough to get sufficient amount of light at the exit slit for easy observation.)

Additional note: This last equation for $\Delta \lambda$ comes from $\Delta \theta=\frac{\Delta x}{f}$ and using $m \lambda=d(sin(\theta)+sin(\phi))$ with $m=1$ and taking a derivative which gives $\Delta \lambda=d \Delta \theta$ (approximately). This spread of $\Delta \lambda$ is actually what is usually observed from a spectrometer. If the slits are made very narrow, then the $\Delta \lambda$ can approach the diffraction limit given by the first formula for $\Delta \lambda$ in the above post. $\\$ $\\$ This first formula for $\Delta \lambda$ is derived by using the equation for a grating of $N$ groves $I(\theta)=I_o \frac{sin^2(N \phi/2)}{sin^2(\phi/2)}$ where $\phi=\frac{2 \pi}{\lambda}d ( \sin(\theta)+sin(\phi))$ and analyzing this function to determine the resolution. This is just some additional info for you, and it should be sufficient to simply use the formula for the resolving power $R$ provided in the previous post above. $\\$ $\\$ Incidentally, I just worked the first part of this problem and there is some ambiguity in the problem. Depending on how the diffracted angle is defined, there can be a minus sign on the $sin(\phi)$ term in the formula. It does appear for this case that that is what they are doing and the correct answer for the order $m$ may in fact be $m=3$ instead of $m=10$. (The helper isn't normally supposed to provide answers, but here it is necessary to be able to discuss the complete problem.) Also, then in using the formula for $R=Nm$, you would use $m=3$, etc. Also it is not uncommon for a diffraction grating spectrometer to be used with order $m=3$, (most often $m=1$ is used), but generally, most measurements are not performed at an order as high as 9 or 10.(For the diffracted angle, the problem should state which side of the perpendicular to the grating it is defining as positive, i.e. it would do well to supply a diagram. Otherwise, it leaves the person solving the problem to try to guess which direction they are referring to.)

 

[ReidMerrill]

Additional note: This last equation for $\Delta \lambda$ comes from $\Delta \theta=\frac{\Delta x}{f}$ and using $m \lambda=d(sin(\theta)+sin(\phi))$ with $m=1$ and taking a derivative which gives $\Delta \lambda=d \Delta \theta$ (approximately). This spread of $\Delta \lambda$ is actually what is usually observed from a spectrometer. If the slits are made very narrow, then the $\Delta \lambda$ can approach the diffraction limit given by the first formula for $\Delta \lambda$ in the above post. $\\$ $\\$ This first formula for $\Delta \lambda$ is derived by using the equation for a grating of $N$ groves $I(\theta)=I_o \frac{sin^2(N \phi/2)}{sin^2(\phi/2)}$ where $\phi=\frac{2 \pi}{\lambda}d ( \sin(\theta)+sin(\phi))$ and analyzing this function to determine the resolution. This is just some additional info for you, and it should be sufficient to simply use the formula for the resolving power $R$ provided in the previous post above. $\\$ $\\$ Incidentally, I just worked the first part of this problem and there is some ambiguity in the problem. Depending on how the diffracted angle is defined, there can be a minus sign on the $sin(\phi)$ term in the formula. It does appear for this case that that is what they are doing and the correct answer for the order $m$ may in fact be $m=3$ instead of $m=10$. (The helper isn't normally supposed to provide answers, but here it is necessary to be able to discuss the complete problem.) Also, then in using the formula for $R=Nm$, you would use $m=3$, etc. Also it is not uncommon for a diffraction grating spectrometer to be used with order $m=3$, (most often $m=1$ is used), but generally, most measurements are not performed at an order as high as 9 or 10.(For the diffracted angle, the problem should state which side of the perpendicular to the grating it is defining as positive, i.e. it would do well to supply a diagram. Otherwise, it leaves the person solving the problem to try to guess which direction they are referring to.)

Thank you! My book only had half of the resolving power formula.