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(A) The path difference $\delta$ at a point on the screen directly in front of one of the slits is given by $\delta = \frac{b^2}{2d}$.For destructive

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$\frac{b^2}{d}, \frac{b^2}{3d}$

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(A) The path difference $\delta$ at a point on the screen directly in front of one of the slits is given by $\delta = \frac{b^2}{2d}$.For destructive interference (missing wavelengths),the path difference must be an odd multiple of $\frac{\lambda}{2}$,i.e.,$\delta = (2n-1)\frac{\lambda}{2}$ where $n = 1, 2, 3, ...$.Setting $\delta = \frac{b^2}{2d}$,we get $\frac{b^2}{2d} = (2n-1)\frac{\lambda}{2}$.For $n=1$,$\frac{b^2}{2d} = \frac{\lambda_1}{2} \implies \lambda_1 = \frac{b^2}{d}$.For $n=2$,$\frac{b^2}{2d} = \frac{3\lambda_2}{2} \implies \lambda_2 = \frac{b^2}{3d}$.Thus,the missing wavelengths are $\frac{b^2}{d}$ and $\frac{b^2}{3d}$.

White light is used to illuminate two slits in Young's double-slit experiment. The separation between the slits is $b$ and the screen is at a distance $d (d >> b)$ from the slits. The wavelengths missing at a point on the screen directly in front of one of the slits are

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