In a Young's double-slit experiment,white lig | vedclass

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(C) The path difference $\Delta x$ at the point on the screen directly in front of one of the slits is given by $\Delta x = \sqrt{b^2 + d^2} - d$.Sinc

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$1, 3$

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(C) The path difference $\Delta x$ at the point on the screen directly in front of one of the slits is given by $\Delta x = \sqrt{b^2 + d^2} - d$.Since $d >> b$,we can use the binomial approximation: $\sqrt{b^2 + d^2} = d(1 + b^2/d^2)^{1/2} \approx d(1 + b^2/2d^2) = d + b^2/2d$.Thus,$\Delta x \approx (d + b^2/2d) - d = b^2/2d$.For destructive interference (missing wavelengths),the path difference must be an odd multiple of $\lambda/2$:$\Delta x = (2n - 1)\lambda/2$,where $n = 1, 2, 3, ...$Equating the two expressions for $\Delta x$:$b^2/2d = (2n - 1)\lambda/2$$\lambda = b^2 / [(2n - 1)d]$.For $n = 1$,$\lambda = b^2/d$ (Option $1$).For $n = 2$,$\lambda = b^2/3d$ (Option $3$).For $n = 3$,$\lambda = b^2/5d$.Therefore,the missing wavelengths are $b^2/d$ and $b^2/3d$,which corresponds to options $1$ and $3$.

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