Question

The yellow light from a sodium vapor lamp seems to be of pure wavelength, but it produces two first-order maxima at$\mathbf{36}\mathbf{.}{\mathbf{093}}^{\mathbf{°}}$and $\mathbf{36}\mathbf{.}{\mathbf{129}}^{\mathbf{°}}$when projected on a $\mathbf{10}\mathbf{,}\mathbf{000}$line per centimeter diffraction grating. What are the two wavelengths to an accuracy of $\mathbf{0}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{nm}$?

Short answer

The sodium vapor lamp has two wavelengths,$589.1\mathrm{nm}\mathrm{and}589.6\mathrm{nm}$.

Step-by-step solution

Given data

Grating lines on the diffraction grating are 10000 liner per cm, which means that the

distance between the two slits is$\mathrm{d}=\left(\frac{1\mathrm{cm}}{10000}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=1\times {10}^{‐6}\hspace{0.33em}\mathrm{m}$

The order of maximum is 1.

The angles at which the first-order maximum is observed $36.{093}^{°}\mathrm{and}36.{129}^{°}$

Evaluating the wavelengths

The equation of the constructive interference can be expressed as

$\mathbf{\theta sin\lambda }$ …………………(1)

Where, d, $\mathrm{\lambda }$, $\mathrm{\theta }$and m is the separation between slits, the wavelength of the incident light, and the angle made with the maximum and order of maximum, respectively.

Therefore, if we substitute ${\mathrm{\theta }}_1-36.{093}^{°}$and m=1 in equation (1), we will get,

$$\begin{gathered} {\mathrm{\lambda }}_1=\frac{{\mathrm{\theta sin}}_1}{\mathrm{m}} \\ =\frac{\left(1\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(36.{093}^{°}\right)}{1} \\ =589.1\mathrm{nm} \end{gathered}$$

And, if we substitute ${\mathrm{\theta }}_2=36.{129}^{°}$and m=1 in equation (1), we will get,

role="math" localid="1654055552622" $$\begin{gathered} {\mathrm{\lambda }}_2=\frac{\mathrm{d}{\mathrm{sin\theta }}_2}{\mathrm{m}} \\ =\frac{\left(1\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(36.{129}^{°}\right)}{1} \\ =589.6\mathrm{nm} \end{gathered}$$

Therefore, the wavelengths are: $589.1\mathrm{nm}\mathrm{and}589.6\mathrm{nm}$..