Question

Derive Eq. 36-28, the expression for the half-width of the lines in a grating’s diffraction pattern

Short answer

The required equation for the half width of the lines is diffraction grating pattern is$∆\theta =\frac{\lambda }{Nd\cos \theta }$ .

Step-by-step solution

Step 1; Write the given data from the question.

The equation 36.28 from the textbook is

$∆\theta =\frac{\lambda }{Nd\cos \theta }$

Determine the expression to derive the equation 36.28 for the half-width of the lines in a grating’s diffraction pattern.

The expression to calculate the phase between the two adjacent slit is given as follows.

$\mathbf{∆}\mathit{\varphi }\mathbf{=}\mathbf{2}\mathbf{\pi m}\mathbf{+}\frac{\mathbf{2}\mathbf{\pi }}{\mathbf{N}}$

Here, $\mathit{N}$is the number of the slits and localid="1663142714174" $\mathit{m}$is the order of diffraction.

The expression to calculate the path difference between the two adjacent slit is given as follows.

…… (i)

Here, $\mathit{\lambda }$is the wavelength.

Derive the equation 36.28 for the half-width of the lines in a grating’s diffraction pattern.

The condition for the diffraction maxima is given by,

$d\sin \theta =m\lambda$

Calculate the path difference,

Substitute$\left(2\mathrm{\pi m}+\frac{2\mathrm{\pi }}{\mathrm{N}}\right)$for $∆\varphi$in to equation (i).

…… (ii)

$$\begin{gathered} ∆L=\left(2\mathrm{\pi m}+\frac{2\mathrm{\pi }}{\mathrm{N}}\right)\frac{\lambda }{2\mathrm{\pi }} \\ ∆L=2\mathrm{\pi }\left(m+\frac{1}{\mathrm{N}}\right)\frac{\lambda }{2\mathrm{\pi }} \\ ∆\mathrm{L}=\left(\mathrm{m}+\frac{1}{\mathrm{N}}\right)\mathrm{\lambda } \\ ∆\mathrm{L}=\mathrm{m\lambda }+\frac{\mathrm{\lambda }}{\mathrm{N}} \end{gathered}$$

Let assume the$\theta$is the angular position of the $m^{th}$order maxima, and $d$is the slit separation, therefore the path difference is also given by,

…… (iii)

$∆L=sdin(\theta +∆\theta ]$

Equate the equation (ii) and (iii).

$\begin{array}{ll}m\lambda +\frac{\lambda }{N}=d\sin \left(\theta +∆\theta \right)& \\ m\lambda +\frac{\lambda }{N}=d\left[\sin \left(\theta \right)\cos \left(∆\theta \right)+\sin \left(∆\theta \right)\cos \left(\theta \right)\right]& \end{array}$

Since the is very small, therefore $\sin \left(∆\theta \right)=∆\theta$and$\cos \left(∆\theta \right)=1$ .

$\begin{array}{l}m\lambda +\frac{\lambda }{N}=d(sin\theta \times 1+∆\theta \cos \theta )\\ m\lambda +\frac{\lambda }{N}=dsin\theta +d∆\theta cos\theta \end{array}$

Substitute$m\lambda$ for $d\sin \mathrm{\theta }$into above equation.

role="math" localid="1663142681080" $$\begin{gathered} m\lambda +\frac{\lambda }{n}=m\lambda +d∆\theta \cos \theta \\ \frac{\lambda }{n}=d∆\theta \cos \theta \\ ∆\theta =\frac{\lambda }{Nd\cos \theta } \end{gathered}$$

Hence the required equation for the half width of the lines is diffraction grating pattern is$∆\theta =\frac{\lambda }{Nd\cos \theta }$ .