91Ó°ÊÓ

The maxima of the diffraction gratings are given by $\mathrm{m}=\frac{\mathrm{λ}}{\mathrm{»å²õ¾\pm ²Ôθ}}$

Let the distance between the diffraction grating and the screen to be $D$and the distance between the central axis and any point on the screen to be $y$.

So, the angle at the inner edge of the square can be obtained as follows:

$$\begin{gathered} \tan \mathrm{θ}=\frac{y}{D} \\ \mathrm{θ}={\tan }^{-1}\left(\frac{y}{D}\right) \end{gathered}$$

Substitute $y=50×{10}^{-3}\text{m}$and $D=0.300 \text{m}$:

$$\begin{gathered} \mathrm{θ}={\tan }^{-1}\left(\frac{50×{10}^{-3}}{0.300}\right) \\ =9.46° \end{gathered}$$

Here, the diffraction grating is at the rate of$350 \text{lines/mm}$. So, the value of$d$for this case can be obtained as follows:

$$\begin{gathered} d=\frac{1}{350} \\ =2.86×{10}^{-6} \text{m} \end{gathered}$$

Substitute$\mathrm{θ}=9.46°$and $d=2.86×{10}^{-6} \text{m}$in $m=\frac{d\sin \mathrm{θ}}{\mathrm{λ}}$:

$$\begin{gathered} m=\frac{2.86×{10}^{-6}\sin \left(9.46°\right)}{\mathrm{λ}} \\ m=\frac{470}{\mathrm{λ}} \text{nm} \end{gathered}$$

Here, the wight light the wavelength is $\mathrm{λ}>400 \text{nm}$, So, the only allowed value of$m$is$m=1$. So,

$$\begin{gathered} 1=\frac{470}{\mathrm{λ}} \\ \mathrm{λ}=470 \end{gathered}$$

Thus, the shortest wavelengths of the light that pass through the hole is$470nm$.

(b)

At the outer edge, the value of angle can be obtained as follows:

$$\begin{gathered} \tan \mathrm{θ}=\frac{y+10×{10}^{-3}}{D} \\ \mathrm{θ}={\tan }^{-1}\left(\frac{50×{10}^{-3}+10×{10}^{-3}}{0.300}\right) \\ \mathrm{θ}=11.31° \end{gathered}$$

Substitute $\mathrm{θ}=11.31°$and$d=2.86×{10}^{-6} \text{m}$in$m=\frac{d\sin \mathrm{θ}}{\mathrm{λ}}$:

$$\begin{gathered} m=\frac{2.86×{10}^{-6}\sin \left(11.31°\right)}{\mathrm{λ}} \\ m=\frac{560}{\mathrm{λ}} \text{nm} \end{gathered}$$

Here, the wight light the wavelength is$\mathrm{λ}>400 \text{nm}$, So, the only allowed value of $m$is$m=1$. So,

$$\begin{gathered} 1=\frac{560}{\mathrm{λ}} \\ \mathrm{λ}=560 \text{nm} \end{gathered}$$

Thus, the longest wavelengths of the light that passes through the hole is$470nm$ .