Physics - Wave Optics Question with Solution | TestHub

The position of the $n$-th secondary maximum in a single slit diffraction pattern is given by $y_n=\left(n+\frac{1}{2}\right)\frac{\lambda D}{a}$.

For the first secondary maximum, $n=1$, so $y_1=\frac{3}{2}\frac{\lambda D}{a}$.

The separation between the positions of the first secondary maximum for the two wavelengths is:

$\Delta y=\frac{3}{2}\frac{D}{a}({\lambda }_2-{\lambda }_1)$

Given $D=2\text{ m}$, $a=2.4\times 10^{-3}\text{ m}$, ${\lambda }_1=590\times 10^{-9}\text{ m}$, and ${\lambda }_2=596\times 10^{-9}\text{ m}$.

Substituting the values:

$\Delta y=\frac{3}{2}\times \frac{2}{2.4\times 10^{-3}}\times (596-590)\times 10^{-9}$

$\Delta y=\frac{3}{2.4\times 10^{-3}}\times 6\times 10^{-9}$

$\Delta y=\frac{18}{2.4}\times 10^{-6}$

$\Delta y=7.5\times 10^{-6}\text{ m}$

Answer: $7.5\times 10^{-6}\mathrm{m}$