Width of one of the two slits in a Young's double slit interference experiment is half of the other slit.…

Width of one of the two slits in a Young's double slit interference experiment is half of the other slit. The ratio of the maximum to the minimum intensity in the interference pattern is :
  1. $(2 \sqrt{2}+1):(2 \sqrt{2}-1)$
  2. $(3+2 \sqrt{2}):(3-2 \sqrt{2})$
  3. $9: 1$
  4. $3: 1$

Solution

$\mathrm{I} \propto$ width $\quad\quad \mathrm{I}_{\max }=\left(\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}\right)^2$
$\therefore \mathrm{I}_1=\mathrm{I}_0, \mathrm{I}_2=2 \mathrm{I}_0 \quad\quad \mathrm{I}_{\min }=\left(\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}\right)^2$
$\frac{I_{\max }}{I_{\min }}=\frac{(\sqrt{2}+1)^2}{(\sqrt{2}-1)^2} \Rightarrow \frac{3+2 \sqrt{2}}{3-2 \sqrt{2}}$

Asked in: JEE Main 2025 (03 Apr Shift 2)

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