In Young's double slits experiment, the light has a frequency \( 12 \times 10^{14}\,\mathrm{Hz} \) and the distance between the centers of adjacent fringes is 0.8 mm. If the screen is 1.6 m away, what is the distance between the slits? (Take speed of light \( 3 \times 10^{8}\,\mathrm{m\,s^{-1}} \))
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 5.00 \times 10^{-4}\,\mathrm{m} \)
The distance between the centres of adjacent fringes is the fringe width \( \beta = \lambda D/d \). The wavelength comes from the frequency through \( c = \nu\lambda \).
\( \displaystyle \lambda = \frac{c}{\nu} = \frac{3\times10^{8}\,\mathrm{m\,s^{-1}}}{12\times10^{14}\,\mathrm{Hz}} = 2.5\times10^{-7}\,\mathrm{m} \)
\( \beta = 0.8\,\mathrm{mm} = 8\times10^{-4}\,\mathrm{m} \)
\( D = 1.6\,\mathrm{m} \)
\( \displaystyle d = \frac{\lambda D}{\beta} = \frac{2.5\times10^{-7} \times 1.6}{8\times10^{-4}}\,\mathrm{m} \)
\( \displaystyle d = \frac{4\times10^{-7}}{8\times10^{-4}}\,\mathrm{m} = 5.00\times10^{-4}\,\mathrm{m} \)