Q61 / 75 · JEE Main 2025 (23 Jan Shift 2) · Physics

The width of one of the two slits in Young's double slit experiment is d while that of the other slit is $x \mathrm{~d}$. If the ratio of the maximum to the minimum intensity in the interference pattern on the screen is $9: 4$ then what is the value of $x$ ? (Assume that the field strength varies according to the slit width.)

  • A
    4
  • Correct answer
    B
    5
  • C
    3
  • D
    2

Solution

$\begin{aligned} & \mathrm{I} \propto(\text { width })^2 \\ & \left(\frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}\right)^2=\frac{9}{4} \\ & \frac{\sqrt{\mathrm{I}_1}+\sqrt{\mathrm{I}_2}}{\sqrt{\mathrm{I}_1}-\sqrt{\mathrm{I}_2}}=\frac{3}{2} \\ & \frac{(\mathrm{x}+1) \mathrm{d}}{(\mathrm{x}-1) \mathrm{d}}=\frac{3}{2} \\ & \Rightarrow 3 \mathrm{x}-3=2 \mathrm{x}+2 \\ & \mathrm{x}=5\end{aligned}$

Hence, the correct answer is option: 'B'.

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