In a Young's double slit experiment, the wavelength of light used is 600 nm, separation between the slits is 0.1 mm, and the distance between the slits and the screen is 1.0 m. The fringe width is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 6 mm
In Young’s double slit experiment the fringe width is \( \beta = \lambda D/d \), with every length in metres.
\( \lambda = 600\,\mathrm{nm} = 6\times10^{-7}\,\mathrm{m} \)
\( d = 0.1\,\mathrm{mm} = 1\times10^{-4}\,\mathrm{m} \)
\( D = 1.0\,\mathrm{m} \)
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{(6\times10^{-7}\,\mathrm{m}) \times (1.0\,\mathrm{m})}{1\times10^{-4}\,\mathrm{m}} \)
\( \beta = 6\times10^{-3}\,\mathrm{m} = 6\,\mathrm{mm} \)