Question

At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?

Short answer

The angle for the first-order maximum of the blue-light of 450-nm is ${}^{0.52°}$.

Step-by-step solution

Given data

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

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

Finding the angle which is the first-order maximum

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

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

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

\(\begin{array}{l}(5 \times {10^{ - 5}}\;m) \times \sin \theta = 4.5 \times {10^{ - 7}}\;m\\\theta = {\sin ^{ - 1}}\left( {\frac{{4.5 \times {{10}^{ - 7}}\;{\rm{m}}}}{{5 \times {{10}^{ - 5}}\;{\rm{m}}}}} \right)\\\theta = 0.52^\circ \end{array}\)

Therefore, the angle is${}^{0.52°}$for the first-order maximum.