Monochromatic light is passed through a double slits, with separation between the slits, 0.5 mm and interference pattern is observed on a screen 2.5 m away. The first bright fringe is formed 2 mm away from central fringe. What is the wavelength of light?
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 400 nm
In Young’s double-slit experiment the n-th bright fringe lies at \( y_n = n\lambda D/d \) from the central fringe, where D is the slit-screen distance and d the slit separation. For the first bright fringe n = 1.
\( d = 0.5\,\mathrm{mm} = 0.5\times10^{-3}\,\mathrm{m},\quad y_1 = 2\,\mathrm{mm} = 2\times10^{-3}\,\mathrm{m},\quad D = 2.5\,\mathrm{m} \)
\( \displaystyle \lambda = \frac{y_1 d}{D} = \frac{(2\times10^{-3})(0.5\times10^{-3})}{2.5} \)
\( \lambda = 4\times10^{-7}\,\mathrm{m} = 400\,\mathrm{nm} \)
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 – …
