Question

Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of 45.0º.

Short answer

The distance between two slits that produces the first minimum is ${}^{2.9\text{'}{10}^{-7}}$.

Step-by-step solution

Given data

The wavelength of the violet light is ${}^{\lambda =}410\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=4.1\times {10}^{-7}\mathrm{m}$

The angle for the first maximum is ${}^{45.0°}$

Finding the separation between two slits

A formula for the angle of the first-order minimum can be expressed as,

$\mathrm{d}\mathrm{sin\theta }=\frac{1}{2}\mathrm{\lambda }$………………………….(1)

Substituting the given data in equation (1), we get,

$$\begin{gathered} d\times \sin \left(45.0°\right)=\frac{1}{2}\times 4.1\times {10}^{-7}m \\ d\times \sin \left(45°\right)=2.05\times {10}^{-7}m \\ d=\left(\frac{2.05\times {10}^{-7}m}{0.707}\right) \\ d=2.9\times {10}^{-7}m \end{gathered}$$

Thus, the distance between two slits is ${}^{2.9\text{'}{10}^{-7}\mathrm{m}}$.