In a Young's double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen is Io where the path difference between two interfering waves is λ . The path difference between the interfering waves at a point on the screen where the intensity is Io/4 will be-
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → λ/3
In Young’s experiment with two equal slits, the resultant intensity depends on the phase difference \( \phi \) between the waves:
\( \displaystyle I = I_{\mathrm{max}}\cos^{2}\frac{\phi}{2}, \quad \phi = \frac{2\pi}{\lambda}\,\Delta x \)
For a path difference \( \Delta x = \lambda \), \( \phi = 2\pi \) and \( \cos^{2}\pi = 1 \), so Io is the maximum intensity. Now put \( I = I_o/4 \):
\( \displaystyle \frac{I_o}{4} = I_o\cos^{2}\frac{\phi}{2} \Rightarrow \cos\frac{\phi}{2} = \frac{1}{2} \)
\( \displaystyle \frac{\phi}{2} = \frac{\pi}{3} \Rightarrow \phi = \frac{2\pi}{3} \)
\( \displaystyle \Delta x = \frac{\lambda}{2\pi}\,\phi = \frac{\lambda}{2\pi}\times\frac{2\pi}{3} = \frac{\lambda}{3} \)