Question

At what angle is the fourth-order maximum for the situation in Exercise 27.6? 13.

Short answer

The angle for the fourth-order maximum is${}^{2.{06}^0}$.

Step-by-step solution

Given data

The wavelength of the violet light is $\mathrm{\lambda }=450\mathrm{nm}\left(\frac{{10}^{-9}\mathrm{m}}{1\mathrm{nm}}\right)=4.5\times {10}^{-7}\mathrm{m}$

The separation between the two slits is$\mathrm{d}=0.0500\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=5\times {10}^{-5}\mathrm{m}$

Finding the angle

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

$\mathrm{dsin\theta }=4\mathrm{\lambda }.........................................\left(1\right)$

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

$(5\times {10}^{-5}\mathrm{m})\times \mathrm{sin\theta }=4\times 4.5\times {10}^{-7}\mathrm{m}$

$\theta ={\sin }^{-1}\left(\frac{18.0\times {10}^{-1}\mathrm{m}}{5\times {10}^{-5}\mathrm{m}}\right)$

$\mathrm{\theta }=20.6^0$

Therefore, the angle of maximal fourth-order is ${}^{2.{06}^0}$