Question

Some water waves with a wavelength of 3.0 m are diffracted by an array of evenly spaced posts in the water. If the rows of posts are separated by 5.0 m, calculate the angle $\mathbf{2}\mathit{\theta }$at which the first-order “Bragg diffraction” of these water waves will be seen.

Short answer

The value of $2\theta =34.8^{\circ }$.

Step-by-step solution

Bragg’s law

Bragg’s diffraction occurs whenelectromagnetic radiationwaves have wavelengths comparable to atomic spacing in a crystal lattice.

X-ray diffraction is used for the identification of crystals.

$\mathbf{n\lambda }\mathbf{=}\mathbf{2}\mathbf{d} \mathbf{sin\theta }$is Bragg’s equation.

Calculation

Given,

The distance between the row of posts$d=5.0 \text{m}$.

The wavelength of the water waves is 3.0 m.

For the first order Bragg’s diffraction, n=1.

Hence, the value of $\theta$is calculated as shown below:

$\begin{array}{c}n\lambda =2d \sin \theta \\ 1\times 3.0 m=2\times 5.0 m \times \sin \theta \\ \sin \theta =\frac{3}{10}\\ \theta ={\sin }^{-1}\left(0.3\right)\\ \theta =17.4\end{array}$

As $\theta =17.4°$

$\begin{array}{c}2\theta =2\times 17.4°\\ =34.8°\end{array}$

Hence, $2\theta =34.8°$.