Question

A grating has 350 rulings/mm and is illuminated at normal incidence by white light. A spectrum is formed on a screen 30.0 cm from the grating. If a hole 10.0 mm square is cut in the screen, its inner edge being 50.0 mm from the central maximum and parallel to it, what are the (a) shortest and (b) longest wavelengths of the light that passes through the hole?

Short answer

1. The shortest wavelengths of the light that passes through the hole is$470nm$ .
2. The longest wavelengths of the light that passes through the hole is$560nm$ .

Step-by-step solution

Maxima of Diffraction

The maxima of the diffraction gratings are given by $\mathrm{m}=\frac{\mathrm{\lambda }}{\mathrm{dsin\theta }}$

Solution of part (a)

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:

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

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

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

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:

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

Substitute$\theta =9.46°$and $d=2.86\times {10}^{-6} \text{m}$in $m=\frac{d\sin \theta }{\lambda }$:

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

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

$$\begin{gathered} 1=\frac{470}{\lambda } \\ \lambda =470 \end{gathered}$$

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

Solution of part (b)

(b)

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

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

Substitute $\theta =11.31°$and$d=2.86\times {10}^{-6} \text{m}$in$m=\frac{d\sin \theta }{\lambda }$:

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

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

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

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