Question

What is the smallest separation between two slits that will produce a second-order maximum for ${}^{\mathbf{720}\mathbf{-}\mathbf{nm}}$ red light?

Short answer

The smallest separation between two slits that produces a second-order maximum is obtained as ${}^{1.44\times {10}^{-6}\mathrm{m}}$.

Step-by-step solution

Given data

The wavelength of the light is$\mathrm{\lambda }=720\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=7.20\times {10}^{-7}\mathrm{m}$

The order of the maximum is 2.

Evaluating the smallest separation

A formula for the angle of the second-order maximum can be expressed as,

$\mathrm{dsin\theta }=2\mathrm{\lambda }.................................\left(1\right)$

We are then supposed to solve for the smallest value of d using the fact the smallest value of d corresponds to${}^{\mathrm{\theta }={90}^0}$.

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

$$\begin{gathered} \mathrm{d}=\frac{2\mathrm{\lambda }}{\mathrm{sin\theta }} \\ =\frac{2\times 7.20\times {10}^{-7}\mathrm{m}}{\sin \left({90}^0\right)} \\ =1.44\times {10}^{-6}\mathrm{m} \end{gathered}$$

Therefore, the smallest separation is${}^{1.44\times {10}^{-6}\mathrm{m}}$.