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(A) In a Young's double slit experiment,the position of the central maximum is determined by the condition that the path difference between the waves

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Vertically downwards slightly

Vedclass · Answer

(A) In a Young's double slit experiment,the position of the central maximum is determined by the condition that the path difference between the waves from the two slits is zero. Let the path difference be $\Delta x = (S_B + B P) - (S_A + A P) = 0$. Initially,the path difference is zero at the center of the screen. When the optical path length from $S$ to $B$ is increased,the path difference at the original central point becomes $\Delta x = (S_B + \Delta S_B + B P) - (S_A + A P) = \Delta S_B > 0$. To maintain the condition $\Delta x = 0$,the point $P$ must shift such that the path length from $A$ becomes greater than the path length from $B$. Since slit $A$ is above slit $B$,the central maximum shifts towards the slit with the increased optical path length,which is slit $B$. Therefore,the fringe system shifts vertically downwards.

Similar Questions

Light of wavelength $600 \, nm$ is incident on a double slit and the interference fringes are formed on a screen $1 \, m$ apart. The separation between two consecutive dark fringes on a screen is found to be $1.2 \, mm$. What is the separation between the slits in $mm$?

In a Young's double-slit experiment,the slits are separated by $0.12 \, mm$ and the screen is at a distance of $1 \, m$. Find the distance of the $3^{rd}$ dark fringe from the center of the screen in $cm$. Given $\lambda = 6000 \, \mathring{A}$.

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