Question

Find the angle for the third-order maximum for ${}^{\mathbf{580}\mathbf{-}\mathbf{nm}}$-wavelength yellow light falling on a diffraction grating having ${}^{\mathbf{1500}}$ lines per centimeter.

Short answer

The third-order maximum is obtained at an angle of ${}^{\mathbf{15}\mathbf{.}{\mathbf{1}}^{\mathbf{0}}}$.

Step-by-step solution

Given data

Grating lines on the diffraction grating are 1500 liner per cm, which means that the distance between the two slits is $\mathrm{d}=\left(\frac{1\mathrm{cm}}{1500}\right)\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=6.67\times {10}^{-6}\mathrm{m}$

The wavelength of the yellow light is $\mathrm{\lambda }=580\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=5.80\times {10}^{-7}\mathrm{m}$

The order of maximum is 3.

Evaluating the third-order maximum

For the given value of ${}^{\mathrm{\theta }}$, and ${}^{\mathrm{\lambda }}$, the equation for the third-order maximum can be expressed as$\mathrm{d}\mathrm{sin\theta }=3\mathrm{\lambda }$,

The value of ${}^{\mathrm{\theta }}$can be estimated from the above expression as,

$$\begin{gathered} \mathrm{\theta }={\sin }^{-1}\left(\frac{\mathrm{\lambda }}{\mathrm{d}}\right) \\ ={\sin }^{-1}\left(\frac{3\times 5.8\times {10}^{-7}\mathrm{m}}{6.67\times {10}^{-6}\mathrm{m}}\right) \\ ={\sin }^{-1}(0.261) \\ =15.1^0 \end{gathered}$$

Therefore, the third-order maximum is, ${}^{15.1^0}$.