Chapter 47: Problem 22

Question: A grating has $620~{\rm rulings/mm}$ and is $w~=~5.05~{\rm mm}$wide. (a) What is the smallest wavelength interval that can be resolved in the third order at $\lambda~=~481~{\rm nm}$. (b) How many higher orders can be seen?

Solution: The adjacent slit seperation is

$$d~=~{10^{-3}\over 620}~=~1.61\times 10^{-6}~{\rm m}$$

In order to resolve two wavelengths we require that

$$\Delta\lambda~=~{\lambda\over N m}$$

where m is the order and N is the total number of lines on the grating. Hence

$$\Delta\lambda~=~{\lambda\over N m}~=~{481\over 5.05 620 3}~=~0.051~{\rm nm}$$

Since

$$d\sin\theta_k~=~k\lambda$$

and $\sin\theta~=~1$ for the maximum order, we have that $k_{\rm max}~=~3.34$, and hence the third order is the highest that can be seen.