In a Young’s double slit experiment, \( I' \) is the intensity at a point, where the path difference between the interfering waves is \( \displaystyle\frac{\lambda}{6} \). If \( I_{0} \) denotes the maximum intensity observed in the interference pattern, then the ratio of \( I'/I_{0} \) is: [Given: \( \lambda \) is the wavelength of light]
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( \displaystyle\frac{3}{4} \)
The phase difference corresponding to a path difference \( \Delta x \) is \( \phi = (2\pi/\lambda)\Delta x \). For two slits of equal intensity, the resultant intensity is \( I = I_0\cos^{2}(\phi/2) \), where \( I_0 \) is the maximum intensity.
\( \displaystyle \phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{6} = \frac{\pi}{3} \)
\( \displaystyle \frac{I'}{I_0} = \cos^{2}\frac{\phi}{2} = \cos^{2}\frac{\pi}{6} = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} \)
Note: A similar question was also asked in CUET (UG) 2023 on 05-06-2023 (Slot 1).