In a Young’s double slit experiment, the separation between the slits is \( 0.2 \) mm and the distance of the screen from the slits is \( 25 \) cm. If the wavelength of the light used is \( 500 \) nm, what is the fringe width?
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( 0.625\,\mathrm{mm} \)
The fringe width in Young’s double slit experiment is \( \beta = \lambda D/d \).
\( \lambda = 500\,\mathrm{nm} = 5\times10^{-7}\,\mathrm{m} \)
\( D = 25\,\mathrm{cm} = 0.25\,\mathrm{m} \)
\( d = 0.2\,\mathrm{mm} = 2\times10^{-4}\,\mathrm{m} \)
\( \displaystyle \beta = \frac{\lambda D}{d} = \frac{(5\times10^{-7})(0.25)}{2\times10^{-4}}\,\mathrm{m} = 6.25\times10^{-4}\,\mathrm{m} = 0.625\,\mathrm{mm} \)