The intensity at a point A on the screen in Young’s double-slit experiment is half the intensity of the central bright fringe. The path difference between the two waves at A is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \lambda/4 \)
In Young’s experiment the two slits act as coherent sources of equal intensity \( I_0 \). At a point where the phase difference is \( \phi \), the intensity is \( I=4I_0\cos^2(\phi/2) \). At the central bright fringe \( \phi=0 \), so the intensity there is \( 4I_0 \).
At A the intensity is half of this:
\( \displaystyle 4I_0\cos^2\frac{\phi}{2}=\frac{1}{2}\times4I_0 \)
\( \displaystyle \cos^2\frac{\phi}{2}=\frac{1}{2} \)
\( \Rightarrow\; \phi=\pi/2 \)
A path difference of \( \lambda \) gives a phase difference of \( 2\pi \), so the path difference at A is:
\( \begin{aligned} \Delta &= \frac{\lambda}{2\pi}\,\phi \\ &= \frac{\lambda}{2\pi}\times\frac{\pi}{2} \\ &= \frac{\lambda}{4} \end{aligned} \)