In the interference pattern of a Young's double slit experiment, the phase difference between the coherent waves, which produce 3rd maximum is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6π
In Young’s double slit experiment a bright fringe (maximum) is formed where the path difference is a whole number of wavelengths. The central bright fringe is the zeroth maximum (n = 0), so the 3rd maximum has n = 3.
\( \displaystyle \Delta x = n\lambda \)
A path difference of one wavelength corresponds to a phase difference of \( 2\pi \).
\( \displaystyle \phi = \frac{2\pi}{\lambda}\,\Delta x = 2\pi n \)
\( \displaystyle \phi = 2\pi \times 3 = 6\pi \)