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(A) In Young's double slit experiment, the path difference $\Delta x$ at a point $y$ on the screen is given by $\Delta x = \frac{yd}{D}$.For a point e

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$d^2/D, d^2/3D, d^2/5D, \dots$

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(A) In Young's double slit experiment, the path difference $\Delta x$ at a point $y$ on the screen is given by $\Delta x = \frac{yd}{D}$.For a point exactly in front of one slit, $y = d/2$.Therefore, the path difference is $\Delta x = \frac{(d/2)d}{D} = \frac{d^2}{2D}$.For a minimum (destructive interference) to occur, the path difference must be an odd multiple of $\lambda/2$, i.e.,$\Delta x = (2n-1)\frac{\lambda}{2}$ where $n = 1, 2, 3, \dots$.Equating the two expressions: $\frac{d^2}{2D} = (2n-1)\frac{\lambda}{2}$.Solving for $\lambda$: $\lambda = \frac{d^2}{D(2n-1)}$.Thus, $\lambda \propto \frac{1}{2n-1}$.For $n=1, 2, 3, \dots$, the values of $(2n-1)$ are $1, 3, 5, \dots$.Therefore, the possible wavelengths are proportional to $\frac{d^2}{D}, \frac{d^2}{3D}, \frac{d^2}{5D}, \dots$.

In Young's double slit experiment, in an interference pattern, a minimum is observed exactly in front of one slit. The distance between the two coherent sources is $d$ and $D$ is the distance between the source and the screen. The possible wavelengths used are inversely proportional to

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