Physics - Wave Optics Question with Solution | TestHub

The position of the $m^{\text{th}}$ maximum in Young's double slit experiment is given by $y=\frac{m\lambda D}{d}$.

For the first case, the distance of the $n^{\text{th}}$ maximum from the central maximum is:

$y_1=\frac{n{\lambda }_{1}D}{d}$

For the second case, the distance of the ${\left(\frac{n}{3}\right)}^{\text{th}}$ maximum from the central maximum is:

$y_2=\frac{\left(\frac{n}{3}\right){\lambda }_{2}D}{d}$

Taking the ratio of $y_1$ and $y_2$:

$\frac{y_1}{y_2}=\frac{\frac{n{\lambda }_{1}D}{d}}{\frac{n{\lambda }_{2}D}{3d}}=\frac{3{\lambda }_1}{{\lambda }_2}$

Answer: $\frac{3{\lambda }_1}{{\lambda }_2}$