In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda /3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda /6$. The ratio of intensities at points $A$ and $B$ is _______.

[JEE Main 2026, 6 Apr (Shift 1)]

Question

In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda /3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda /6$. The ratio of intensities at points $A$ and $B$ is _______.

[JEE Main 2026, 6 Apr (Shift 1)]

YESPYQ · Accepted Answer

Correct answer: c) $\frac{1}{3}$. (c) The phase difference $\phi$ is related to the path difference $\Delta x$ by: $ϕ=\frac{2\pi }{\lambda }\Delta x$ Path difference $\Delta {x}_{A}=\frac{\lambda }{3}$ ${ϕ}_{A}=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}\$ ${I}_{A}={I}_{0}{\cos }^{2}\left(\frac{2\pi /3}{2}\right)={I}_{0}{\left(\frac{1}{2}\right)}^{2}=\frac{{I}_{0}}{4}$ Path difference $\overset{¨}{A}{x}_{B}=\frac{\lambda }{6}$ ${ϕ}_{B}=\frac{2\pi }{\lambda }\times \frac{\lambda }{6}=\frac{\pi }{3}\$ ${I}_{B}={I}_{0}{\cos }^{2}\left(\frac{\pi /3}{2}\right)={I}_{0}{\cos }^{2}\left(\frac{\pi }{6}\right)\$ $={I}_{0}{\left(\frac{\sqrt{3}}{2}\right)}^{2}=\frac{3{I}_{0}}{4}$ Ratio of intensities: $\frac{{I}_{A}}{{I}_{B}}=\frac{{I}_{0}/4}{3{I}_{0}/4}=\frac{1}{3}$

In interference experiment the path difference between two interfering waves at a point \(A\) on the screen is \(\lambda…

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