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For example, when $a\sin\theta=3\lambda$, we rewrite this as $\frac{a}{6}\sin\theta=\frac{\lambda}{2}$ and explain that the waves from every pair of point source whose path difference is $\frac{a}{6}\sin\theta$ will destructively interfere, leading to a dark fringe.

Since each element of area of the slit opening can be considered as a source of secondary waves according to Huygen's principle, what if we divided the slit into five strips of equal width, and say that two point sources, whose path difference $\frac{a}{5}\sin\theta$ is equal to $\frac{\lambda}{2}$ will destructively interfere to form a dark fringe, meaning that $a\sin\theta=\frac{5\lambda}{2}$. What is the reason behind why we can't do this?

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    $\begingroup$ I think in your first paragraph is a mistake. If they interfere destructively it should give a dark fringe, not a bright fringe. $\endgroup$ Commented Jul 18, 2021 at 13:26
  • $\begingroup$ @CharlesTucker3 Thank you, edited post $\endgroup$
    – Freddie
    Commented Jul 19, 2021 at 9:27

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Yes. You may divide the slit into 5 segments. Each segment has a $\lambda /2 $ optical path difference to the next segment, as shown in the following figure.

enter image description here

Thus, the contribution of segment A (to the diffraction amplitude) cancels that of segment B, segment C cancels segment D. But, you still have a contribution from E left.

Therefore, $d \sin\theta = \frac{5}{2} \lambda$ is not a condition for dark fringe.

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The logic behind these separation into pairs of rays that interfere destructively is not to separate the single slit into 5 or 6 pieces, but into an infinite number, and each element produces a wave according to Huygen's principle, as you correctly said. For some angles, fulfilling the criterion of destructive interference $d \cdot \sin(\alpha) = m \cdot \lambda$, we can find pairs of these infinitely small (and infinitely many) pieces so that each secondary wave cancels out with that from the partner.

Note, the arguments here are for the far-field, i.e. Fraunhofer diffraction. In this approximantion, for $\sin \alpha = \frac{\lambda}{d}$ (first minimum) the path difference between the uppermost ray and the central ray (when hitting the screen which is in infinite distance...) is $\frac{\lambda}{2}$, and thus they interfere destructively. The ray immediately below the uppermost ray has also a partner, which is the ray immediately below the central ray (their path difference is also $\frac{\pi}{2}$). In direction $\alpha$, both rays will interfere destrcutively. And so on... So each small segment has a partner and cancels out with it.

For the second minimum, $\sin(\alpha) = \frac{2\lambda}{d}$, we can also find a partner for each segment. Now, the path difference between the uppermost ray and the ray a quarter of $d$ below the upermost ray cancels out. And rays immediately below them form a pair. The central rays's partner for destructive interference is the ray at $\frac{3d}{4}$. I think you get the idea...

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