Two coherent plane waves having wavelength \(\lambda\) and wavevectors
\[ \begin{aligned} \vec{k}_{1} &= \frac{2\pi}{\lambda}(\frac{1}{2}\hat{\imath} - \frac{\sqrt{3}}{2}\hat{\jmath}) \text{ and} \\ \vec{k}_{2} &= \frac{2\pi}{\lambda}(-\frac{1}{2}\hat{\imath} - \frac{\sqrt{3}}{2}\hat{\jmath}), \end{aligned} \]
are incident on a screen placed on the \(xz\) - plane. The overlap of the waves on the plane of the screen produces interference fringes.
The fringe width (center-to-center spacing of bright fringes) will be:
IIT JAM - 2026,
Answer: D
The fringe width is \(\lambda\).
Two plane waves interfere with intensity \(\propto 1 + \cos(\Delta\vec{k} \cdot \vec{r})\), where \(\Delta\vec{k} = \vec{k}_1 - \vec{k}_2\). The \(\hat{\jmath}\) parts are equal and cancel:
\[ \begin{aligned} \Delta\vec{k} &= \frac{2\pi}{\lambda}\left(\tfrac{1}{2} + \tfrac{1}{2}\right)\hat{\imath} \\ &= \frac{2\pi}{\lambda}\hat{\imath} \end{aligned} \]
On the screen (the \(xz\)-plane) the phase difference is \(\frac{2\pi}{\lambda}x\), so bright fringes repeat every time \(x\) increases by \(\lambda\). Equivalently, each wave makes 30° with the screen normal, the angle between them is 60°, and the fringe spacing \(\frac{\lambda}{2\sin 30^\circ} = \lambda\).
The \(\hat{\jmath}\) components play no part, since they are identical in both waves and only set the common direction of travel; \(\frac{\lambda}{2}\), \(2\lambda\) and \(\frac{\sqrt{3}\lambda}{2}\) do not match the spacing \(2\pi/|\Delta k_x|\) fixed by the \(\hat{\imath}\) components.
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