Question

When monochromatic light is incident on a slit 22.0 $\mathbf{\mu }$m wide, the first diffraction minimum lies at 1.80° from the direction of the incident light. What is the wavelength?

Short answer

The wavelength is 691 nm.

Step-by-step solution

Describe the diffraction by a single slit results in maxima

The expression to calculate the wavelength is given by,

$$\begin{gathered} \mathit{m}\mathit{\lambda }\mathbf{=}\mathit{d}\mathit{s}\mathit{i}\mathit{n}\mathit{\theta } \\ \mathit{\lambda }\mathbf{=}\frac{\mathbf{d}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{\theta }}{\mathbf{m}} \end{gathered}$$ ….. (1)

Here, $\mathit{\lambda }$is the wavelength, $\mathit{d}$is separation, and $\mathit{\theta }$is angle.

Determine the wavelength

Substitute $22\times {10}^{-3}\text{m}$for $d$, $1.80°$for $\theta$,and 1 for $m$in equation (1).

$$\begin{gathered} \lambda =\frac{\left(22\times {10}^{-3}\text{m}\right)\sin \left(1.80°\right)}{1} \\ =6.91\times {10}^{-7}\text{m} \\ =691\times {10}^{-9}\text{m} \\ =691\text{nm} \end{gathered}$$

Therefore, the wavelength is 691 nm.