Question

Calculate the wavelength of light that has its third minimum at an angle of 30.0º when falling on double slits separated by 3.00 µm.

Short answer

The wavelength of the light is${}^{600\mathrm{nm}}$.

Step-by-step solution

Given data

The angle for the third minimum is ${}^{30.0°}$

The separation of the double slits is${}^{d=3.00\mu m\left(\frac{{10}^{-°}}{1\mu m}\right)=3.00\times {10}^{-6}m}$

Finding the wavelength of the light

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

$\mathrm{d}\sin \mathrm{\theta }=\frac{5}{2}\mathrm{\lambda }$………………………….(1)

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

$$\begin{gathered} 3.00\times {10}^{-6}m\times \sin \left(30.0°\right)=\frac{5}{2}\times \lambda \\ \lambda =\left(\frac{1.50\times {10}^{-6}m}{2.5}\right) \\ \lambda =0.600\times {10}^{-6}m\left(\frac{1nm}{{10}^{-9}m}\right) \\ \lambda =600nm \end{gathered}$$

Thus, the distance between two slits is ${}^{600\mathrm{nm}}$.