In a Young’s double slit experiment, the slits are separated by \( 0.28 \) mm and the screen is placed \( 1.4 \) m away from the slit. The distance between the central bright fringe and the fourth bright fringe is measured to be \( 1.4 \) cm. The wavelength of the light used is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 700 nm
In Young’s double-slit experiment the \( n \)th bright fringe lies at \( y_{n} = n\lambda D/d \) from the central fringe, so the fourth bright fringe is four fringe widths away.
\( d = 0.28\,\mathrm{mm} = 2.8\times10^{-4}\,\mathrm{m},\ D = 1.4\,\mathrm{m},\ y_{4} = 1.4\,\mathrm{cm} = 1.4\times10^{-2}\,\mathrm{m} \)
\( \displaystyle y_{4} = \frac{4\lambda D}{d} \Rightarrow \lambda = \frac{y_{4}d}{4D} \)
\( \displaystyle \lambda = \frac{(1.4\times10^{-2}\,\mathrm{m})(2.8\times10^{-4}\,\mathrm{m})}{4\times1.4\,\mathrm{m}} \)
\( \lambda = 7\times10^{-7}\,\mathrm{m} = 700\,\mathrm{nm} \)
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 – …
