Question

Sodium light is incident normally on a grating of width 4×10−3
m. The total number
of lines on the grating is 2000. Determine the angular separation between the sodium
D-lines in the first order spectrum.

Answer 1

Answer on Question #81239 Physics / Optics

Question. Sodium light is incident normally on a grating of width $4 \cdot 10^{-3} \, \text{m}$. The total number of lines on the grating is 2000. Determine the angular separation between the sodium $D$-lines in the first order spectrum?

Solution.

This light contains two closely spaced lines (the well-known sodium doublet) of wavelengths $589 \, \text{nm}$ and $589.59 \, \text{nm}$. The grating spacing $d$ is given by

$$d = \frac{l}{N} = \frac{4 \cdot 10^{-3}}{2000} = 2 \cdot 10^{-6} \, \text{m}.$$

For first line

$$\begin{array}{l} d \cdot \sin \theta_1 = m \lambda_1 \rightarrow \theta_1 = \arcsin \left(\frac{m \lambda_1}{d}\right) = \\ = \arcsin \left(\frac{1 \cdot 589 \cdot 10^{-9}}{2 \cdot 10^{-6}}\right) = 0.298932 \, \text{rad} = 17.13{}^\circ. \end{array}$$

For second line

$$\begin{array}{l} d \cdot \sin \theta_2 = m \lambda_2 \rightarrow \theta_2 = \arcsin \left(\frac{m \lambda_2}{d}\right) = \\ = \arcsin \left(\frac{1 \cdot 589.59 \cdot 10^{-9}}{2 \cdot 10^{-6}}\right) = 0.299241 \, \text{rad} = 17.15{}^\circ. \end{array}$$

The angular separation is

$$\Delta \theta = 0.299241 - 0.298932 = 0.000309 \, \text{rad} = 1'4''.$$

Answer. $\Delta \theta = 0.000309 \, \text{rad} = 1'4''$.

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