A diffraction grating has resolving power $\mathit{R}\mathbf{=}\frac{{\mathbf{\lambda }}_{\mathbf{avg}}}{\mathbf{\Delta \lambda }}\mathbf{=}\mathit{N}\mathit{m}$. (a) Show that the corresponding frequency range f that can just be resolved is given by $\mathbf{∆}\mathit{f}\mathbf{=}\frac{\mathbf{c}}{\mathbf{Nm\lambda }}$. (b) From Fig. 36-22, show that the times required for light to travel along the ray at the bottom of the figure and the ray at the top differ by $\mathbf{∆}\mathit{t}\mathbf{=}\mathbf{\left(}\frac{\mathbf{Nd}}{\mathbf{C}}\mathbf{\right)}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }$. (c) Show that $\mathbf{\left(}\mathbf{\Delta f}\mathbf{\right)}\mathbf{\left(}\mathbf{\Delta t}\mathbf{\right)}$, this relation being independent of the various grating parameters. Assume $\mathit{N}\mathbf{\gg }\mathbf{1}$.

The given data can be listed below,

• The diffracting grating is,$R=\frac{{\lambda }_{avg}}{\Delta \lambda }=Nm$
• The frequency range is,$\Delta f=\frac{c}{Nm\lambda }$

A diffraction grating is an optical element with a prism-like effect that divides white light into its individual hues. To divide light into portions based on wavelength, a diffraction grating functions similarly to a prism. The incident light angle is distorted by its tiny, typically periodic characteristics.

(a)

The wavelength range that can only be resolved in order m is given by,

$\Delta \lambda =\frac{\lambda }{Nm}$ ………(i)

Here, N is the grating's number of rulings, and$\lambda$ stands for the average wavelength.

The relationship between frequency and wavelength is given by,

$f=\frac{c}{\lambda }$ ……..… (ii)

Here, c is the speed of light,

Consequently,

$f\Delta \lambda +\lambda \Delta f=0$ ……….(iii)

The change in the wavelength is given by,

$$\begin{gathered} \Delta \lambda =-\frac{\lambda }{f}\Delta f \\ =-\frac{{\lambda }^2}{c}\Delta f .......... \left(\text{iv}\right) \end{gathered}$$

Equate equation (i) and (iv) change in frequency is given by,

$$\begin{gathered} \frac{\lambda }{Nm}=-\frac{{\lambda }^2}{c}\Delta f \\ \Delta f=\frac{c}{\lambda Nm} \end{gathered}$$

Thus, it is proved that corresponding frequency range f that can just be resolved is given by$\Delta f=\frac{c}{Nm\lambda }$.

(b)

When waves are going along the two extreme rays, the difference in travel time is given by,

$\Delta t=\frac{\Delta L}{c}$

Here, L is the route length difference.

The path difference is given by

$\Delta L=\left(N-1\right)d\sin \theta$,

Since the waves begin at slits that are separated by (N - 1)d.

Substitute all the values in the above and N is large, the time difference is given by

$$\begin{gathered} \left(N-1\right)d\sin \theta =c\Delta t \\ \Delta t=\frac{\left(N-1\right)d\sin \theta }{c} \end{gathered}$$

If the Nis large the time difference is given by,

$\Delta t=\frac{Nd}{c}\sin \theta$

Thus, the lens does not affect the travel time, so it is proved the times required for light to travel along the ray at the bottom of the figure and the ray at the top differ by

$\Delta t=\frac{Nd}{c}\sin \theta$

(c)

Brag’s law of diffraction is given by,

$d\sin \theta =m\lambda$

Here, dis the intermolecular distance,$\lambda$ is the wavelength.

The change in frequency and time is given by,

$\Delta f=\frac{c}{\lambda Nm}$

The time difference is given by,

$\Delta t=\frac{Nd}{c}\sin \theta$

The product of the frequency and time change is given by,

$$\begin{gathered} \Delta f\Delta t=\frac{c}{\lambda Nm}\frac{Nd}{c}\sin \theta \\ =\frac{d\sin \theta }{m\lambda } \\ =\frac{m\lambda }{m\lambda } \\ =1 \end{gathered}$$

Thus, it is proved that the multiplication of frequency and time change is constant so, it is independent of diffraction grating parameters.