Question

A diffraction grating 20.0 mm wide has 6000 rulings. Light of wavelength 589 nm is incident perpendicularly on the grating. What are the

(a) largest,

(b) second largest,

(c) third largest values of $\mathbf{\theta }$at which maxima appear on a distant viewing screen?

Short answer

(a)$62.1°$

(b)$45.0°$

(c)$32.0°$

Step-by-step solution

Identification of the given data

The given data is listed below as-

• The width of the diffraction grating is 20.0 mm.
• The number of rulings = 6000.
• The wavelength of light, $\lambda =589\text{nm}$ .

The condition of the diffraction grating

The condition of the diffraction grating is:

$dsin\theta =m\lambda$

Here, d is the distance between adjacent rulings and $\lambda$ is the wavelength of light.

To find the largest values ofθat which maxima appear on a distant viewing screen (a)

The distance between adjacent rulings is:

$$\begin{gathered} d=20.006000 \\ =0.00333mm \\ =3.33\mu m \end{gathered}$$

Now, the condition of the diffraction grating is:

$d\sin \theta =m\lambda$

Where, $m=0,\pm 1,\pm 2,....$

Now, the largest value of $\theta$corresponds to $\left|m\right|=5$.

Therefore, the value of $\theta$is:

$\begin{array}{l}\theta ={\sin }^{-1}\left(\left|m\right|\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$d$}\right.\right)\\ ={\sin }^{-1}\left(\frac{5(0.589)\mu m)}{3.33\mu \mathrm{m}}\right)\\ =62.1^0\end{array}$

Thus,the largest value of$\theta$at which maxima appear on a distant viewing screen is $62.1°$.

To find the second largest values of θ at which maxima appear on a distant viewing screen(b)

The second largest value of $\theta$corresponds to $\left|m\right|=4$.

Therefore, the value of $\theta$is:

$\begin{array}{l}\theta ={\sin }^{-1}\left(\left|m\right|\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$d$}\right.\right)\\ ={\sin }^{-1}\left(\frac{4(0.589)\mu m)}{3.33\mu \mathrm{m}}\right)\\ =45.0^{\circ }\end{array}$

Thus,the second largest value of $\theta$ at which maxima appear on a distant viewing screen is $45.0°$.

To find the third largest values of θ at which maxima appear on a distant viewing screen(c)

The third largest value of $\theta$ corresponds to$\left|m\right|=3$ .

Therefore, the value of $\theta$ is:

role="math" localid="1663159335558" $\begin{array}{l}\theta ={\sin }^{-1}\left(\left|m\right|\raisebox{1ex}{$\lambda $}\!\left/ \!\raisebox{-1ex}{$d$}\right.\right)\\ ={\sin }^{-1}\left(\frac{3(0.589)\mu m)}{3.33\mu \mathrm{m}}\right)\\ =32.0^{\circ }\end{array}$

Thus, the third largest value of $\theta$ at which maxima appear on a distant viewing screen is $32°$.