Question 287: Waves Optics

Question

The phase difference between two waves giving rise to dark fringe in Young's double slit experiment is ( n is the integer)

Accepted Answer

In Young's double-slit experiment, the phase difference between the two waves is related to the path difference by
$$\Delta \phi = \frac{2\pi}{\lambda} \Delta r.$$
For a dark fringe (destructive interference), the path difference must be
$$\Delta r = \left(n + \frac{1}{2}\right) \lambda,$$
where $$n$$ is an integer (starting from 0). Substituting this into the phase difference formula gives
$$\Delta \phi = \frac{2\pi}{\lambda} \left(n + \frac{1}{2}\right) \lambda = 2\pi \left(n + \frac{1}{2}\right) = (2n+1)\pi.$$
Notice that writing $(2n+1)\pi$ or $(2n-1)\pi$ essentially represents the odd multiples of $$\pi$$; the difference lies only in the indexing of $$n$$ (whether you start counting from 0 or 1). In this problem, Option C is written as
$$(2n-1)\pi,$$
which is equivalent to the dark fringe condition if we assume $$n$$ starts from 1.
Thus, the correct answer is Option C: $$(2n-1)\pi.$$

Answer

zero

Answer

$(4 \mathrm{n}+1) \frac{\pi}{2}$

Answer

$(2 n-1) \pi$

Answer

$(2 n+1) \frac{\pi}{2}$

Waves Optics

The phase difference between two waves giving rise to dark fringe in Young's double slit experiment is ( n is the integer)

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