# Resolving Power of a collimated Littrow diffracting telescope

Tags:
1. Mar 12, 2013

### alphaparrot

1. The problem statement, all variables and given/known data
Suppose a simplified telescope and grating spectrometer system:

D is the telescope's primary diameter, W is the width of the collimated beam, β is the grating angle in a Littrow configuration, and $\alpha$ is the aperture diameter in angular units on the sky.

The goal is to find an expression of the resolving power R that depends only on those 4 quantities. We can assume that the detector size in the exit focal plane is small compared to the projected aperture size.

2. Relevant equations

R=$\frac{\lambda}{\Delta\lambda}$
R=mN where N is the number of grating slits or ridges and m is the diffraction order.

Since the grating is in Littrow configuration, we know that 2dsinβ=mλ, so λ=$\frac{2d\sin{\beta}}{m}$ where d is the distance between ridges/slits.

D and W are related to $\alpha$ by D=$2f_1\sin{\alpha}$ and W=$2f_2\sin{\alpha}$, where $f_1$ and $f_2$ are the focal lengths from the primary lens to the aperture and from the collimating lens to the aperture, respectively.

We know the number of slits/ridges illuminated is given by N=$\frac{W}{d\cos{\beta}}$

If I'm not mistaken, the reflected light is normal to the grating.

3. The attempt at a solution

$\frac{\delta\lambda}{\delta\beta}=\frac{2d\cos{β}}{m}$

Therefore, approximating $\delta\lambda$ to Δλ, R=$\frac{\tan{\beta}}{\delta\beta}$

So from this point, we need $\delta\beta$ in terms of $\alpha$, D, and W. But here is where I'm stuck. I can't for the life of me figure out what to do next.