In Young’s double-slit experiment, light of wavelength 600 nm is used. At a point P on the screen, the path difference between the two waves is 1.8 μm. How many dark fringes are there between the central bright fringe and P?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 3
A dark fringe forms where the path difference is \( (n+\frac{1}{2})\lambda \). The path difference grows from zero at the central bright fringe to its value at P.
\( \lambda=600\ \mathrm{nm}=0.6\ \mu\mathrm{m} \)
\( \begin{aligned} \frac{\Delta x}{\lambda} &= \frac{1.8}{0.6}=3 \\ \Delta x &= 3\lambda \end{aligned} \)
So P is a bright fringe with \( n=3 \). Between zero and \( 3\lambda \), the dark fringes are at \( 0.5\lambda,\ 1.5\lambda \) and \( 2.5\lambda \). Hence, there are 3 dark fringes between the central bright fringe and P.
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.”