Question

A 4.0-cm-wide diffraction grating has 2000 slits. It is illuminated by light of wavelength $550\mathrm{nm}$. What are the angles (in degrees) of the first two diffraction orders?

Short answer

The angle in degree of the first two diffraction order is${1.6}^{\circ }\text{and}{3.2}^{\circ }$

Step-by-step solution

Find angle(part a)

We know that the formula for a diffraction grating connecting the m-t h diffraction angle, the order m of the diffraction, the wavelength, and the space between two consecutive gratings is

$d\sin {\theta }_m=m\lambda$

As a result, the angle at which m-t h order diffraction is observed is

$\theta ={\sin }^{-1}\left(\frac{m\lambda }{d}\right)$

Find angle(part b)

Knowing that there are 2000 gratings in$4\mathrm{cm}$, the grating constant will be $500\mathrm{l}/\mathrm{cm}$. This means that the separation d is $d=\frac{1\mathrm{cm}}{500}=2·{10}^{-5}\mathrm{m}$.

In our case, $\frac{\lambda }{d}=\frac{5.5·{10}^{-7}}{2·{10}^{-5}}=0.0275<<1$, which means we can apply the angle approximation to find the angle as ${\theta }_m=\frac{m\lambda }{d}$. We will then, for our angles, have${\theta }_1=0.025\mathrm{rad}=0.0275\cdot \frac{{360}^{\circ }}{2\pi }={1.6}^{\circ }$

${\theta }_1=0.025\mathrm{rad}=0.0275\cdot \frac{{360}^{\circ }}{2\pi }={1.6}^{\circ }$

${\theta }_2=0.055\mathrm{rad}=0.055\cdot \frac{{360}^{\circ }}{2\pi }={3.2}^{\circ }$