Question

What is the wavelength of light falling on double slits separated by 2.00 µm if the third-order maximum is at an angle of 60.0º ?

Short answer

The wavelength of the light is${}^{577\mathrm{nm}}$.

Step-by-step solution

Given data

The angle for the third minimum is${}^{\mathbf{60}\mathbf{.}{\mathbf{0}}^{\mathbf{0}}}$

The separation of the double slits is$\mathrm{d}=2.00\mathrm{\mu m}\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{\mu m}}\right)=2.00\times {10}^{-6}\mathrm{m}$

Finding the wavelength of the light

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

$\mathrm{dsin\theta }=3\mathrm{\lambda }...............................\left(1\right)$

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

$$\begin{gathered} 2.00\times {10}^{-6}\mathrm{m}\times \sin \left(60.0^0\right)=3\times \mathrm{\lambda } \\ \mathrm{\lambda }=\left(\frac{1.732\times {10}^{-6}\mathrm{m}}{3}\right) \\ \mathrm{\lambda }=0.577\times {10}^{-6}\mathrm{m}\left(\frac{1\mathrm{nm}}{{10}^{-9}\mathrm{m}}\right) \\ \mathrm{\lambda }=577\mathrm{nm} \end{gathered}$$

Thus, the distance between two slits is ${}^{\mathbf{577}\mathbf{}\mathbf{nm}}$.