Question

Show that a diffraction grating cannot produce a second-order maximum for a given wavelength of light unless the first-order maximum is at an angle less than$\mathbf{30}\mathbf{.}\mathbf{0}\mathbf{°}$

Short answer

The diffraction grating can’t produce a second-order maximum for a wavelength of light if the first-order maximum is obtained beyond $30.0^{°}$.

Step-by-step solution

Given data

The maximum angle for the first-order maximum is$30.0°$.

Evaluating the diffraction that cannot produce second-order maximum

The equation of the constructive interference can be expressed as

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

Where, d,$\lambda$,$\mathrm{\theta }$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 (role="math" localid="1654069179134" $\mathrm{m}=1$) and role="math" localid="1654069123332" ${\mathrm{\theta }}_2$is the angle for the second-order maximum ($\mathrm{m}=2$).

Therefore using equation (1) for both angles will give,

$$\begin{gathered} {\mathrm{\theta sin}}_1=\frac{\mathrm{\lambda }}{\mathrm{d}} \\ {\mathrm{\theta sin}}_2=\frac{2}{\mathrm{d}} \\ \frac{{\mathrm{\theta sin}}_1}{{\mathrm{\theta sin}}_2}=\frac{1}{2} \end{gathered}$$

The maximum value for the angle of diffraction will be ${\mathrm{\theta }}_2=90.0°$.

$$\begin{gathered} {\mathrm{\theta }}_1={\sin }^{‐1}\left(\frac{1}{2}\right)\times \left(\sin \left(90.0°\right)\right) \\ {\mathrm{\theta }}_1={\sin }^{‐1}\left(\frac{1}{2}\right) \\ {\mathrm{\theta }}_1=30.0° \end{gathered}$$

Therefore, the diffraction grating can’t produce a second-order maximum for a wavelength of light if the first-order maximum is obtained beyond $30.0°$.