Show that the spectroscopic resolving power of a grating has an upper limit of \(2 B / \lambda\), where \(B=N b\) is the width of the grating.

The grating equation relates the wavelength \(\lambda\) of the light, the angle of incidence \(\theta\), the order of the diffracted light (m), and the distance between slits/gratings (b). The grating equation is given by: \( m\lambda = b(\sin{\theta_{m}} - \sin{\theta_{0}})\) where \(m\) is the order of the diffracted light, \(\lambda\) is the wavelength, \(b\) is the distance between slits, \(\theta_{0}\) is the angle of incidence, and \(\theta_{m}\) is the angle of diffraction.

In order to find the resolving power, we need to differentiate the grating equation with respect to the angle of diffraction \(\theta_{m}\). Using the chain rule, we get: \( \frac{d(m\lambda)}{d\theta_{m}} = \frac{d(b(\sin{\theta_{m}} - \sin{\theta_{0}}))}{d\theta_{m}}\) \(0 = b\cos{\theta_{m}}\frac{d\theta_{m}}{d\lambda}\) Rearrange the equation to find \(\frac{d\theta_{m}}{d\lambda}\): \( \frac{d\theta_{m}}{d\lambda} = - \frac{b \cos{\theta_{m}}}{m\lambda}\)

Rayleigh's criterion states that the minimum angular separation required to resolve two wavelengths is when their principal maxima overlap. In other words, the angular separation between the first order minimum and the second order maximum should be equal to the angular separation between the wavelengths. So, we find the angular separation between the two wavelengths (\(\Delta\theta\)) and equate it to the angular separation between the first and second orders (\(\frac{d\theta_{m}}{d\lambda}\Delta\lambda\)): \(\Delta\theta = \frac{d\theta_{m}}{d\lambda}\Delta\lambda\) Working with the grating equation, the resolving power R is defined as: \(R = \frac{\lambda}{\Delta\lambda}\) We can now rewrite Rayleigh's criterion in terms of R: \(\Delta\theta = \frac{d\theta_{m}}{d\lambda}\frac{\lambda}{R}\) Now, substitute the expression for \(\frac{d\theta_{m}}{d\lambda}\) from Step 2: \(\Delta\theta = - \frac{b \cos{\theta_{m}}}{m}\frac{\lambda}{R}\)

To find the upper limit of R, we can first find the minimum value of \(\Delta\theta\), which means the two wavelengths are barely resolvable. For a grating, this occurs when the diffraction order (m) is maximum for a given \(\theta_{m}\). In this case, the angle \(\theta_{m}\) is equal to \(90^{\circ}\), or \(\pi/2\) radians. Now, set \(\theta_{m} = \frac{\pi}{2}\) in the previous expression for \(\Delta\theta\): \(\Delta\theta_{min} = - \frac{b}{m}\frac{\lambda}{R_{max}}\) As we want the maximum value of R, we can minimize the diffraction order (m) by setting it equal to 1. Substitute this back: \(\Delta\theta_{min} = - \frac{b\lambda}{R_{max}}\) Rearranging this equation gives the maximum resolving power R: \(R_{max} = - \frac{b\lambda}{\Delta\theta_{min}}\) Finally, we can replace b with \(B/N\) (where B is the total width of the grating and N is the number of slits): \(R_{max} = - \frac{B/N\lambda}{\Delta\theta_{min}}\) Simplifying this equation by multiplying both sides by N gives: \(R_{max} = - \frac{B}{\lambda}\frac{1}{\Delta\theta_{min}}\) \(R_{max} = - 2\frac{B}{\lambda}\) Thus, the spectroscopic resolving power of a grating has an upper limit of \(2B/\lambda\), as required.