Chapter 47: Problem 19

Question: White light ( $400~{\rm nm}\le \lambda\ge 700~{\rm nm}$) is incident on agrating. Show that, no matter what the value of the grating spacing, d, the second- and third-order spectra overlap.

Solution: For a given wavelength, the location of the kth interference fringe is at $d\sin\theta_k~=~k\lambda$. Forming the ratio,r, for two adjacent orders, and two different wavelengths gives

$$r~=~{\sin\theta_2\over\sin\theta_{3}}~=~{2\lambda_2\over 3\lambda_3}$$

If the ratio r is greater than one for any wavelength, then the two spectra overlap. The max value of this ratio is

$$r_{\rm max}~=~{\sin\theta_2\over \sin\theta_{3}}_{\rm max}~=~{2\ 700\over 3\ 400}~=~{7\over 6}$$

hence the spectra of the second and third orders overlap.