In Young’s double-slit experiment, the slits are 0.3 mm apart and the screen is 1.2 m away. Light of wavelength 500 nm is used. The distance of the third dark fringe from the central bright fringe is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 5 mm
Dark fringes are at \( x=(n+\frac{1}{2})\lambda D/d \) with \( n=0,1,2,\ldots \). The first dark fringe has \( n=0 \). So the third dark fringe has \( n=2 \).
\( \lambda=500\ \mathrm{nm}=5\times10^{-7}\ \mathrm{m} \)
\( d=0.3\ \mathrm{mm}=3\times10^{-4}\ \mathrm{m} \)
\( D=1.2\ \mathrm{m} \)
\( \begin{aligned} x &= \left(2+\frac{1}{2}\right)\frac{\lambda D}{d} \\ &= 2.5\times\frac{5\times10^{-7}\times1.2}{3\times10^{-4}} \\ &= 2.5\times2\times10^{-3}\ \mathrm{m} \\ &= 5\ \mathrm{mm} \end{aligned} \)
Hence, the third dark fringe is 5 mm from the central bright fringe.
NCERT (2026-27):
Interference of Light Waves and Young’s Experiment – “On the other hand, we will have destructive interference resulting in a dark region when”
\( \displaystyle \frac{xd}{D}=\left(n+\frac{1}{2}\right)\lambda \)
“that is”
\( \displaystyle x=x_n=\left(n+\frac{1}{2}\right)\frac{\lambda D}{d};\quad n=0,\pm 1,\pm 2 \)\( \qquad (10.14) \)