Question

At what angle does a diffraction grating produces a second-order maximum for light having a first-order maximum at $\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{°}$?

Short answer

The angle of the second-order maximum is obtained as$43.2°$.

Step-by-step solution

Given data

The angle at which the first-order maximum is observed$20.0°$

Evaluating the angle of the second-order maximum

The equation of the constructive interference can be expressed as

$\mathbf{\theta }\mathbf{}\mathbf{sin\lambda }$ …………………(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.

We assume the ${\mathrm{\theta }}_1$is the angle for the first-order maximum ($\mathrm{m}=1$) and ${\mathrm{\theta }}_2$is the angle for the second-order maximum ($\mathrm{m}=2$).

Substituting the given data in equation (1) will give

$$\begin{gathered} {\mathrm{sin\theta }}_1=\frac{{\mathrm{m}}_1\mathrm{\lambda }}{\mathrm{d}}...........................\left(2\right) \\ =\frac{\mathrm{\lambda }}{\mathrm{d}} \\ {\mathrm{sin\theta }}_2=\frac{{\mathrm{m}}_1\mathrm{\lambda }}{\mathrm{d}}...........................\left(3\right) \\ =\frac{2\mathrm{\lambda }}{\mathrm{d}} \end{gathered}$$

Comparing equations (2) and (3)

$$\begin{gathered} \sin \left({\mathrm{\theta }}_2\right)=2\sin \left({\mathrm{\theta }}_1\right) \\ {\mathrm{\theta }}_2={\sin }^{‐1}\left[2\times \left(\sin \left(20.0°\right)\right)\right] \\ {\mathrm{\theta }}_2=43.2° \end{gathered}$$

Therefore, the angle of the second-order maximum is $\mathbf{43}\mathbf{.}\mathbf{2}\mathbf{°}$.