In Young's double slit experiment, the light has a frequency of \( 6\times10^{14}\,\mathrm{Hz} \) and the distance between the centers of adjacent fringes is 0.75 mm. If the screen is 1.5 m away, what is the distance between the slits?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( 10^{-3}\,\mathrm{m} \)
In Young’s double slit experiment the fringe width is \( \beta = \lambda D/d \), where D is the distance of the screen from the slits and d is the slit separation. The wavelength follows from the frequency, \( \lambda = c/\nu \).
\( \displaystyle \lambda = \frac{c}{\nu} = \frac{3\times10^{8}}{6\times10^{14}} = 5\times10^{-7}\,\mathrm{m} \)
\( \displaystyle d = \frac{\lambda D}{\beta} = \frac{5\times10^{-7}\times1.5}{0.75\times10^{-3}} \)
\( d = 1\times10^{-3}\,\mathrm{m} \)