Question

Consider a reflection grating operating with an incident angle of 408 in Figure 20-7. (a) How many lines per centimeter should be etched in the grating if the first-order diffraction angle for 600 nm (visible) light is to be 2308? (b) Answer the same question for 1 000 cm21 (infrared) light.

Short answer

(a)$d=2.38.{10}^{3}lines/cmb)d=142.86lines/cm$

Step-by-step solution

Step: 1 Define the reflection grating:

The light grating has ridges or rulings in its place while the transmission grating has transfer or empty holes in its place.

Step: 2 Grating equation:

(a)

$n.\lambda =d\left(\sin \theta +\sin \varphi \right)$

D is:

$$\begin{gathered} d=\frac{n.\lambda }{\sin \theta +\sin \varphi } \\ d=\frac{1.600.{10}^{-9}m}{\sin \left(40°\right)+\sin \left(-30°\right)} \\ d=4.2.{10}^{-6}m \end{gathered}$$

Or

$$\begin{gathered} d=\frac{1}{4.2.{10}^{-4}cm} \\ d=2.38.{10}^{3}lines/cm \end{gathered}$$

Step: 3 The wavelength:

(b)

$$\begin{gathered} \lambda =\frac{1}{1000c{m}^{-1}} \\ \lambda =1.{10}^{-3}cm \end{gathered}$$

D is:

$$\begin{gathered} d=\frac{n.\lambda }{\sin \theta +\sin \varphi } \\ d=\frac{1.1.{10}^{-3}cm}{\sin \left(40°\right)+\sin \left(-30°\right)} \\ d=7.{10}^{-3}m \end{gathered}$$

Or

$$\begin{gathered} d=\frac{1}{7.{10}^{-3}cm} \\ d=142.86lines/cm \end{gathered}$$

The solution is (a)$d=2.38.{10}^{3}lines/cmb)d=142.86lines/cm$