In Young's double slit experiment the intensity at a point where path difference is \( \lambda/6 \) is I. If \( I_o \) denotes the maximum intensity, then \( I/I_o \) is equal to :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \displaystyle\frac{3}{4} \)
In Young's double slit experiment the intensity at a point depends on the phase difference \( \phi \) between the two waves, \( I = I_o\cos^{2}(\phi/2) \), where \( I_o \) is the maximum intensity (at \( \phi = 0 \)). The phase difference follows from the path difference \( \Delta x \).
\( \displaystyle \phi = \frac{2\pi}{\lambda}\,\Delta x = \frac{2\pi}{\lambda}\times\frac{\lambda}{6} = \frac{\pi}{3} \)
\( \displaystyle \frac{I}{I_o} = \cos^{2}\frac{\phi}{2} = \cos^{2}\frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} \)
Note: A similar question was also asked in CUET (UG) 2026 on 21-05-2026 (Slot 2).