A double slit is illuminated by light of wavelength 600 nm. The slits are 0.1 cm apart and the screen is placed 1 m away. The width of maxima is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 0.6 mm
In Young's double slit experiment the bright fringes (maxima) are equally spaced, and the width of a maximum is the fringe width \( \beta = \lambda D/d \).
\( d = 0.1\,\mathrm{cm} = 1\times10^{-3}\,\mathrm{m},\quad D = 1\,\mathrm{m},\quad \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{6\times10^{-7}\times1}{1\times10^{-3}} = 6\times10^{-4}\,\mathrm{m} = 0.6\,\mathrm{mm} \)
Note: A similar question was also asked in CUET (UG) 2025 on 19-05-2025 (Slot 1).