In Young’s double-slit experiment, the slits are 1.2 mm apart and light of wavelength 600 nm is used. The angular separation between the second bright fringe on one side of the central bright fringe and the third bright fringe on the other side is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(2.5\times10^{-3}\) rad
Bright fringes are at \( x=n\lambda D/d \). Seen from the slits, the angle of a fringe is \( \theta=x/D \). So the \( n \)th bright fringe is at \( \theta_n=n\lambda/d \). The central bright fringe has \( n=0 \).
The two fringes lie on opposite sides of the centre. So their angles add.
\( \lambda=600\ \mathrm{nm}=6\times10^{-7}\ \mathrm{m} \)
\( d=1.2\ \mathrm{mm}=1.2\times10^{-3}\ \mathrm{m} \)
\( \begin{aligned} \Delta\theta &= \theta_2+\theta_3=\frac{5\lambda}{d} \\ &= \frac{5\times6\times10^{-7}}{1.2\times10^{-3}} \\ &= 2.5\times10^{-3}\ \mathrm{rad} \end{aligned} \)
Hence, the angular separation between the two bright fringes is \( 2.5\times10^{-3} \) rad.
NCERT (2026-27):
Interference of Light Waves and Young’s Experiment – “We will have constructive interference resulting in a bright region when”
\( \displaystyle \frac{xd}{D}=n\lambda. \)
“That is,”
\( \displaystyle x=x_n=\frac{n\lambda D}{d};\quad n=0,\pm 1,\pm 2,\ldots \)\( \qquad (10.13) \)