Question

What is the highest-order maximum for 400-nm light falling on double slits separated by 25.0 µm?

Short answer

The Highest-order maximum for 400-nm light falling on double slits separated by 25.0 µm is${}^{62}$.

Step-by-step solution

Given data

The wavelength of the violet light is$\mathrm{\lambda }=400\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=4.0\times {10}^{-7}\mathrm{m}$

The separation between the two slits is$\mathrm{d}=25\mathrm{\mu m}\left(\frac{{10}^{-6}\mathrm{m}}{1\mathrm{nm}}\right)=2.5\times {10}^{-5}\mathrm{m}$

Finding the highest-order maximum

A formula for the angle of the ${\mathrm{m}}^{\mathrm{th}}$-order maximum can be expressed as,

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

The highest order occurs when ${}^{\sin \mathrm{\theta }=1}$.

Therefore, the substitute for these values in equation (1) will give,

$$\begin{gathered} \mathrm{d}=\mathrm{m\lambda } \\ \mathrm{m}=\frac{\mathrm{d}}{\mathrm{\lambda }} \\ \mathrm{m}=\frac{2.5\text{'}{10}^{-5}\mathrm{m}}{4.0\text{'}{10}^{-7}\mathrm{m}} \\ \mathrm{m}=62.5 \end{gathered}$$

Therefore, we will get a maximum order of${}^{\mathrm{m}:62}$