With a particular grating the sodium doublet (589.00 nm and 589.59 nm) is viewed in the third order at $\mathbf{10}\mathbf{°}$ to the normal and is barely resolved. Find (a) the grating spacing and (b) the total width of the rulings.

In diffraction grating, the condition to achieve maxima is given as-

$\mathrm{dsin\theta }=\mathrm{m\lambda }$.

Here, is the separation between grating lines, $\mathrm{\theta }$is the angle made with the normal by the diffracted ray, $\mathrm{m}$ is the order of diffraction and $\mathrm{\lambda }$ is the wavelength.

• Wavelength of the sodium doublet is $589.00\text{nmand}589.59\text{nm}$.
• Angle to the normal is$10°$ .

(a)

Here, the sodium doublet is viewed in the third order at$\theta =10°$ . So,$m=2$ .

Now, the grating spacing can be obtained as follows:

$d=\frac{m{\lambda }_{avg}}{\sin \theta }$

For the given values, the above equation becomes-

$d=\frac{m\left(\frac{589.00+589.59}{2}\right)}{\sin 10°}$

$=\frac{3\left(\frac{589.00\text{nm}+589.59\text{nm}}{2}\right)}{\sin 10°}$

$=1.0\times {10}^4 \text{nm}$

$=10\mu m$

Thus, the grating spacing is$10\mu m$ .

(b)

The width of the ruling $\left(L\right)$can be obtained as the product of the number of rulings $\left(N\right)$and the grating width . The relation can be written as-

$L=N\times d················································\left(1\right)$

The number of rulings on the grating is given as-

$N=\frac{{\lambda }_{avg}}{m\Delta \lambda }$

Here, ${\lambda }_{avg}$is the average wavelength of the doublet and$\Delta \lambda$ is the wavelength separation.

So, the total length $\left(L\right)$ becomes-

$L=\frac{d{\lambda }_{avg}}{m\Delta \lambda }$

For the given values, the above equation becomes-

$L=\frac{\left(10\times {10}^{-6}\right)\text{m}\times \left(\frac{589.00\text{nm}+589.59\text{nm}}{2}\right)}{3\times \left(589.59\text{nm}-589.00\text{nm}\right)}$

role="math" localid="1663060959560" $=3.3\times {10}^3\mu m$

$=3.3mm$

Thus, the total width of the rulings is $3.3mm$.