Two coherent sources vibrate in phase. At a point P, the path difference between their waves is 2.7 μm. For which of the following wavelengths is there destructive interference at P?
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → 600 nm
Destructive interference needs a path difference of a whole number of wavelengths plus half a wavelength. A path difference of a whole number of wavelengths gives a bright point instead.
\( \Delta x=2.7\ \mu\mathrm{m}=2700\ \mathrm{nm} \)
\( \begin{aligned} \Delta x &= \left(n+\frac{1}{2}\right)\lambda \\ \lambda &= \frac{2\Delta x}{2n+1}=\frac{5400\ \mathrm{nm}}{2n+1} \end{aligned} \)
For \( n=0,1,2,\ldots \) this gives 5400 nm, 1800 nm, 1080 nm, 771 nm, 600 nm, 491 nm and so on. Of the given wavelengths, only 600 nm is in this list, with \( n=4 \).
\( \begin{aligned} \frac{\Delta x}{\lambda} &= \frac{2700}{600} \\ &= 4.5=4+\frac{1}{2} \end{aligned} \)
Hence, the interference at P is destructive for light of wavelength 600 nm.
NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “On the other hand, if the point P is such that the path difference,”
\( \displaystyle S_1P\sim S_2P=\left(n+\frac{1}{2}\right)\lambda\quad (n=0,1,2,3,\ldots) \)\( \qquad (10.10) \)
“we will have destructive interference and the resultant intensity will be zero.”