Question

White light (consisting of wavelengths from $\mathbf{400}\mathbf{}\mathit{n}\mathit{m}$ to $\mathbf{700}\mathbf{}\mathit{n}\mathit{m}$) is normally incident on a grating. Show that, no matter what the value of the grating spacing $\mathit{d}$ , the second order and third order overlap.

Short answer

It is proved no matter what the value of the grating spacing $d$, the second order and third order overlap.

Step-by-step solution

Given data                                                                             

White light has wavelengths between

${\lambda }_1=400nm$

${\lambda }_2=700nm$

Definition and concept of diffraction from a grating

An optical element with a periodic structure that divides light into a number of beams that move in diverse directions is known as a diffraction grating.

The angular distance of the $m_{th}$ order diffraction maxima is given by

$\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta }\mathbf{=}\mathit{m}\mathit{\lambda }$ …(i)

Here, $\lambda$is the wavelength of the incident light and $d$ is the separation between two lines in the grating.

Determining the angular distances of second and third order maxima

From equation (i) the angular distance of second order maxima from ${\lambda }_2$ is

$\begin{array}{l}dsin{\theta }_2=2x70\mathrm{mmm}\\ sin{\theta }_2=\frac{1400\mathrm{mm}}{d}\end{array}$

From equation (i) the angular distance of third order maxima from ${\lambda }_1$ is

$\begin{array}{l}dsin{\theta }_3=3x400\mathrm{mm}\\ sin{\theta }_3=\frac{1200\mathrm{mm}}{d}\end{array}$

Thus

$\sin {\theta }_2>\sin {\theta }_3$

Hence the angular distance of a part of the second order from white light is greater than the angular distance of a part of the third order. Thus the second and third order overlap.