JEE Main 2025 — Wave Optics Question with Solution

In Young's double-slit interference experiment, the width of one slit is half that of the other. The intensity, $\mathrm{I}$, is proportional to the slit width. This gives us:

\begin{align*} \mathrm{I}_1 &= \mathrm{I}_0, \\ \mathrm{I}_2 &= 2 \mathrm{I}_0. \end{align*}

The maximum intensity, ${\mathrm{I}}_{\max }$, occurs when the amplitudes add constructively:

${\mathrm{I}}_{\max }={\left(\sqrt{{\mathrm{I}}_1}+\sqrt{{\mathrm{I}}_2}\right)}^2.$

The minimum intensity, ${\mathrm{I}}_{\min }$, occurs when the amplitudes interfere destructively:

${\mathrm{I}}_{\min }={\left(\sqrt{{\mathrm{I}}_1}-\sqrt{{\mathrm{I}}_2}\right)}^2.$

Now, using ${\mathrm{I}}_1={\mathrm{I}}_0$ and ${\mathrm{I}}_2=2{\mathrm{I}}_0$, we find:

$\frac{{\mathrm{I}}_{\max }}{{\mathrm{I}}_{\min }}=\frac{(\sqrt{2}+1{)}^2}{(\sqrt{2}-1{)}^2}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}.$