91Ó°ÊÓ

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, $\mathrm{λ}=589\text{nm}$ .

The condition of the diffraction grating is:

$dsin\mathrm{θ}=m\mathrm{λ}$

Here, d is the distance between adjacent rulings and $\mathrm{λ}$ is the wavelength of light.

The distance between adjacent rulings is:

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

Now, the condition of the diffraction grating is:

$d\sin \mathrm{θ}=m\mathrm{λ}$

Where, $m=0,Â\pm 1,Â\pm 2,....$

Now, the largest value of $\mathrm{θ}$corresponds to $\left|m\right|=5$.

Therefore, the value of $\mathrm{θ}$is:

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

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

The second largest value of $\mathrm{θ}$corresponds to $\left|m\right|=4$.

Therefore, the value of $\mathrm{θ}$is:

$\begin{array}{l}\mathrm{θ}={\sin }^{-1}\left(\left|m\right|\raisebox{1ex}{$\mathrm{λ}$}\!\left/ \!\raisebox{-1ex}{$d$}\right.\right)\\ ={\sin }^{-1}\left(\frac{4(0.589)\mathrm{μ}m)}{3.33\mathrm{μ}\mathrm{m}}\right)\\ =45.0^{∘}\end{array}$

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

The third largest value of $\mathrm{θ}$ corresponds to$\left|m\right|=3$ .

Therefore, the value of $\mathrm{θ}$ is:

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

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