Q 12-10-012NEETJEE MainEasy
In Young's double slit experiment the slit separation is $0.5$ mm, the screen is $1$ m away and the wavelength is $500$ nm. The fringe width is
Answer: (A) $1$ mm
$\beta = \dfrac{\lambda D}{d} = \dfrac{500 \times 10^{-9} \times 1}{0.5 \times 10^{-3}} = 10^{-3}$ m $= 1$ mm.
Solution by Sreeraj P, M.Sc Physics