In a Young's slits experiment the separation of first to fifth fringes is 2.5 mm when the wave length used is 620 nm. The distance from the slits to the screen is 80 cm. Calculate the separation of two slits.

Question:

Solution: The wavelength of light (λ)\(=6.2\times10^{-7}=6.2\times10^{-7}m\),
Distance between the slits and screen (D) = 80 cm = 0.8 m,
Separation of slits (d) =?,
Separation from first to fifth bright fringes (y) = 4β = \(2.5\times10^{-3}\) m 
Here, \(\beta=\frac{\lambda D}d\) is the fringe width,
Now,
\(y=\beta=\frac{4\lambda D}d\)
Or, \(d=\frac{4\lambda D}y\)
Or, \(d=\frac{4\times6.2\times10^{-7}\times0.8}{2.5\times10^{-3}}\)
d = 7.936 x 10^-4 m
Thus, the required distance is, d = 7.936 x 10^-4 m.