In Young’s double slit experiment, two slits are 2 mm apart and fringes are observed on a screen 2 m away from the slits. A light source having two wavelengths, \( 4000\,\mathring{\mathrm{A}} \) and \( 6000\,\mathring{\mathrm{A}} \), is used. The minimum distance from the centre of the screen where the bright fringes produced by both wavelengths coincide is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 1.2 \times 10^{-3}\,\mathrm{m} \)
The bright fringes of the two wavelengths coincide where the \( n \)th fringe of one equals the \( m \)th fringe of the other, \( y = n\lambda_1 D/d = m\lambda_2 D/d \). The nearest coincidence needs the smallest whole numbers in the ratio:
\( \displaystyle n \times 4000 = m \times 6000 \;\Rightarrow\; \frac{n}{m} = \frac{3}{2} \;\Rightarrow\; n = 3,\ m = 2 \)
\( D = 2\,\mathrm{m} \)
\( d = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m} \)
\( \lambda_1 = 4000\,\mathring{\mathrm{A}} = 4\times10^{-7}\,\mathrm{m} \)
\( \displaystyle y = \frac{n\lambda_1 D}{d} = \frac{3 \times 4\times10^{-7} \times 2}{2\times10^{-3}} = 1.2\times10^{-3}\,\mathrm{m} \)
Note: NTA picked this question from an NCERT (2026-27) exercise and changed the figures. Do not ignore NCERT while practising numericals.
NCERT (2026-27):
Exercises – …
