Question

A diffraction grating with a width of 2.0 cm contains 1000 lines/cm across that width. For an incident wavelength of 600 nm, what is the smallest wavelength difference this grating can resolve in the second order?

Short answer

The wavelength difference can resolve in the second order is $\Delta \lambda =0.15 \text{nm}$.

Step-by-step solution

The resolving power

It is known the resolving power of a grating is given by$\mathrm{R}=\mathrm{Nm}=\frac{{\mathrm{\lambda }}_{\mathrm{avg}}}{∆\mathrm{\lambda }}$, where $\mathrm{N}$is the number of rulings in the grating and $\mathrm{m}$is the order of the lines${\mathrm{\lambda }}_{\mathrm{avg}}$, is the average of wavelengths and $∆\mathrm{\lambda }$is the separation.

The wavelength difference

Here, the width of 2.0 cm contains$1000 \text{lines/cm}$ . So, there are total$N=2\times 1000=2000 \text{lines}$ .

Given that the grating is in the second order, so$m=2$ and${\lambda }_{avg}=600 \text{nm}$ . So, the wavelength difference can be obtained as:

$$\begin{gathered} \Delta \lambda =\frac{{\lambda }_{avg}}{Nm} \\ =\frac{600 \text{nm}}{2000\times 2} \\ =0.15 \text{nm} \end{gathered}$$

Thus, the smallest wavelength difference that this grating can resolve in the second order is $\Delta \lambda =0.15 \text{nm}$.