Two identical narrow slits S_1 and S_2 are il | vedclass

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(D) The total path length from the source $P$ to the point $Q$ via slit $S_1$ is $(l_1 + l_3)$.The total path length from the source $P$ to the point

Vedclass · Accepted Answer

$(l_1 + l_3) - (l_2 + l_4) = (2n + 1)\lambda /2$

Vedclass · Answer

(D) The total path length from the source $P$ to the point $Q$ via slit $S_1$ is $(l_1 + l_3)$.The total path length from the source $P$ to the point $Q$ via slit $S_2$ is $(l_2 + l_4)$.The path difference between the two waves reaching point $Q$ is given by $\Delta x = (l_1 + l_3) - (l_2 + l_4)$.For destructive interference to occur at point $Q$,the path difference must be an odd multiple of half the wavelength $\lambda$.Therefore,the condition is $\Delta x = (2n + 1)\lambda / 2$,where $n$ is a positive integer.Substituting the path difference,we get $(l_1 + l_3) - (l_2 + l_4) = (2n + 1)\lambda / 2$.

Two identical narrow slits $S_1$ and $S_2$ are illuminated by light of wavelength $\lambda$ from a point source $P$. If,as shown in the diagram,the light is then allowed to fall on a screen,and if $n$ is a positive integer,the condition for destructive interference at $Q$ is that

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