Question

An electric current through hydrogen gas produces several distinct wavelengths of visible light. What are the wavelengths of the hydrogen spectrum, if they form first-order maxima at angles of${}^{\mathbf{24}\mathbf{.}{\mathbf{2}}^{\mathbf{\circ }}\mathbf{,}\mathbf{25}\mathbf{.}{\mathbf{7}}^{\mathbf{\circ }}\mathbf{,}\mathbf{29}\mathbf{.}{\mathbf{1}}^{\mathbf{\circ }}}$, and role="math" localid="1654239338245" ${}^{\mathbf{41}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}}$when projected on a diffraction grating having ${}^{\mathbf{10}\mathbf{,}\mathbf{000}}$ lines per centimeter? Explicitly show how you follow the steps in Problem-Solving Strategies for Wave Optics.

Short answer

The wavelengths for the first-order maximum observed at angles ${}^{\mathbf{24}\mathbf{.}{\mathbf{2}}^{\mathbf{\circ }}\mathbf{,}\mathbf{25}\mathbf{.}{\mathbf{7}}^{\mathbf{\circ }}\mathbf{,}\mathbf{29}\mathbf{.}{\mathbf{1}}^{\mathbf{\circ }}\mathbf{,}\mathbf{and}\mathbf{}\mathbf{41}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}}$are${}^{\mathbf{410}\mathbf{}\mathbf{nm}\mathbf{,}\mathbf{434}\mathbf{}\mathbf{nm}\mathbf{,}\mathbf{486}\mathbf{}\mathbf{nm}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}\mathbf{656}\mathbf{}\mathbf{nm}}$.

Step-by-step solution

Given data

Grating lines on the diffraction grating are 10000 liner per cm, which means that the $d=\left(\frac{1cm}{10000}\right)\left(\frac{1m}{100cm}\right)=1\times {10}^{‐6}m$

distance between the two slits is

The order of maximum is 1.

The angles at which the first-order maximum is observed ${}^{\mathbf{24}\mathbf{.}{\mathbf{2}}^{\mathbf{\circ }}\mathbf{,}\mathbf{25}\mathbf{.}{\mathbf{7}}^{\mathbf{\circ }}\mathbf{,}\mathbf{29}\mathbf{.}{\mathbf{1}}^{\mathbf{\circ }}\mathbf{,}\mathbf{and}\mathbf{}\mathbf{41}\mathbf{.}{\mathbf{0}}^{\mathbf{\circ }}}$

Evaluating the wavelengths

The equation of the constructive interference can be expressed as

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

Where, d, $\lambda ,\theta$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 ${}^{\mathbf{\theta }\mathbf{=}\mathbf{24}\mathbf{.}{\mathbf{2}}^{\mathbf{\circ }}}$and m=1 in equation (1), we will get,

$$\begin{gathered} \lambda =\frac{\mathrm{dsin\theta }}{\mathrm{m}} \\ =\frac{\left(1.0\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(24.2^{\circ }\right)}{1} \\ =\left(4.1\times {10}^{‐7}\mathrm{m}\right)\left(\frac{1\mathrm{nm}}{{10}^{‐9}\mathrm{m}}\right) \\ =410\mathrm{nm} \end{gathered}$$

For $\theta =25.7^{\circ }$and m=1, the equation (1) will result,

$$\begin{gathered} \lambda =\frac{\mathrm{dsin\theta }}{\mathrm{m}} \\ =\frac{\left(1.0\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(25.7^{\circ }\right)}{1} \\ =\left(4.34\times {10}^{‐7}\mathrm{m}\right)\left(\frac{1\mathrm{nm}}{{10}^{‐9}\mathrm{m}}\right) \end{gathered}$$

$=434\mathrm{nm}$

For $\mathrm{\theta }=29.1^{\circ }$and m=1, the equation (1) will result,

$$\begin{gathered} \lambda =\frac{\mathrm{dsin\theta }}{\mathrm{m}} \\ =\frac{\left(1.0\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(29.1^{\circ }\right)}{1} \\ =\left(4.86\times {10}^{‐7}\mathrm{m}\right)\left(\frac{1\mathrm{nm}}{{10}^{‐9}\mathrm{m}}\right) \\ =486\mathrm{nm} \end{gathered}$$

For $\mathrm{\theta }=41.0^{\circ }$and m=1, the equation (1) will result,

$$\begin{gathered} \lambda =\frac{\mathrm{dsin\theta }}{\mathrm{m}} \\ =\frac{\left(1.0\times {10}^{‐6}\mathrm{m}\right)\times \sin \left(29.1^{\circ }\right)}{1} \\ =\left(6.56\times {10}^{‐7}\mathrm{m}\right)\left(\frac{1\mathrm{nm}}{{10}^{‐9}\mathrm{m}}\right) \\ =656\mathrm{nm} \end{gathered}$$

Therefore, the wavelengths for the first-order maximum observed at angles are.${}^{24.2^{\circ },25.7^{\circ },29.1^{\circ },\mathrm{and}41.0^{\circ }\mathrm{are}434\mathrm{nm},486\mathrm{nm},656\mathrm{nm}.}$