In a Young's double-slit experiment, using monochromatic light of wavelength \( \lambda \), the intensity of light at a point on the screen is \( I_0 \), where path difference between two interfering waves is \( \lambda \). The path difference between the interfering waves at a point where the intensity is \( \displaystyle\frac{I_0}{2} \), will be :
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \lambda/4 \)
For two waves of equal intensity, the resultant intensity depends on the phase difference \( \phi \) as
\( \displaystyle I = I_{\max}\cos^{2}\frac{\phi}{2}, \qquad \phi = \frac{2\pi}{\lambda}\,\Delta x \)
A path difference \( \lambda \) gives \( \phi = 2\pi \) and \( \cos^{2}\pi = 1 \), so that point is a maximum and \( I_{\max} = I_0 \).
For \( I = I_0/2 \):
\( \displaystyle \cos^{2}\frac{\phi}{2} = \frac{1}{2} \;\Rightarrow\; \frac{\phi}{2} = \frac{\pi}{4} \;\Rightarrow\; \phi = \frac{\pi}{2} \)
\( \displaystyle \Delta x = \frac{\lambda}{2\pi}\,\phi = \frac{\lambda}{2\pi}\times\frac{\pi}{2} = \frac{\lambda}{4} \)