In a Young's double-slit experiment, the slits are separated by 0.56 mm and the screen is placed 2.8 m away from the slits. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 600\,\mathrm{nm} \)
The distance from the central maximum to the \( n \)-th bright fringe is \( y_n = n\beta = n\lambda D/d \), where \( \beta \) is the fringe width.
\( \displaystyle y_4 = 4\beta = 1.2\,\mathrm{cm} \Rightarrow \beta = 0.3\,\mathrm{cm} = 3\times10^{-3}\,\mathrm{m} \)
\( \displaystyle \lambda = \frac{\beta d}{D} = \frac{3\times10^{-3} \times 0.56\times10^{-3}}{2.8} = 6\times10^{-7}\,\mathrm{m} = 600\,\mathrm{nm} \)
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 – …
