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(C) In Young's double slit experiment, the condition for destructive interference (dark fringe) is that the path difference $\Delta x$ must be an odd

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$(2n - 1)\pi$

Vedclass · Answer

(C) In Young's double slit experiment, the condition for destructive interference (dark fringe) is that the path difference $\Delta x$ must be an odd multiple of half the wavelength, i.e.,$\Delta x = (2n - 1) \frac{\lambda}{2}$ for $n = 1, 2, 3, \dots$.The relationship between phase difference $\Delta \phi$ and path difference $\Delta x$ is given by $\Delta \phi = \frac{2\pi}{\lambda} \cdot \Delta x$.Substituting the value of $\Delta x$ for the $n^{th}$ dark fringe:$\Delta \phi = \frac{2\pi}{\lambda} \cdot (2n - 1) \frac{\lambda}{2}$.Simplifying the expression:$\Delta \phi = (2n - 1)\pi$ radians.Thus, the correct option is $C$.

In Young's double slit experiment, for the $n^{th}$ dark fringe $(n = 1, 2, 3, \dots)$ the phase difference of the interfering waves in radian will be

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