In Young’s double-slit experiment, the two slits, separated by a distance of 1 mm, are illuminated by a monochromatic light source of wavelength \( 5\times10^{-7}\,\mathrm{m} \). If the distance between the slits and the screen is 2 m, then the separation of the bright fringes in the interference pattern will be
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.0 mm
In Young’s double-slit experiment the bright fringes lie at \( x_n=n\lambda D/d \), so neighbouring bright fringes are separated by the fringe width \( \beta=\lambda D/d \).
With the data in SI units:
\( \lambda=5\times10^{-7}\,\mathrm{m} \)
\( D=2\,\mathrm{m} \)
\( d=1\,\mathrm{mm}=10^{-3}\,\mathrm{m} \)
\( \begin{aligned} \beta &= \frac{\lambda D}{d} \\ &= \frac{5\times10^{-7}\,\mathrm{m}\times2\,\mathrm{m}}{10^{-3}\,\mathrm{m}} \\ &= 10^{-3}\,\mathrm{m} \\ &= 1.0\,\mathrm{mm} \end{aligned} \)