# Grating Spectrometer - Bandpass

Tags:
1. Apr 19, 2014

### unscientific

1. The problem statement, all variables and given/known data

Part (a): Find intensity distribution of N-slit grating
Part (b): Find resolving power
Part (c): Find an expression for Bandpass, and estimate its value. What value of slit width gives same bandpass as resolving power theoretically?

2. Relevant equations

3. The attempt at a solution

I've solved most parts of the question, except the very last part in part (c). I thought the bandpass is independent of wavelength?

Part(a)

$$I = I_0 \frac{sin^2(\frac{Nkd sin \theta}{2})}{sin^2 \frac{kd sin \theta}{2}} sinc^2 (\frac{ka sin \theta}{2})$$

Part (b)

Width from central maxima to first minimum $\Delta \theta = \frac{1}{N}(\frac{\lambda}{d})$

For pth order fringe maximum: $sin \theta_{max} = \frac{p\lambda}{d}$, I get angular dispersion as $\frac{d\theta}{d\lambda} = \frac{p}{d cos \theta} \approx \frac{p}{d}$.

Thus, Resolving Power $\frac{\lambda}{\Delta \lambda} = \frac{\lambda}{\Delta \theta / \frac{d\theta}{d\lambda}} = Np = N$ for p = 1.

Part (c)

Bandpass is defined as the amount of lambda it lets through, so Bandpass is:
$$\frac{d\lambda}{d\theta} \Delta \theta$$
$$= \frac{d\lambda}{d\theta} \frac{w}{f}$$

Where w is exit path width and f is focal length.

I found bandpass = $5.56 * 10^{-5} m$

Now for the last part, theoretically for first order resolving power is independent of λ?

Resolving power = N

$$\frac{wd}{f} = N$$
$$w = 3*10^{11} m$$