Answer:
[tex]1.22 \mu m[/tex]
Explanation:
In the double-slit interference, light passes through a double slit and produce a pattern of alternating bright and dark fringes on a distant screen. This pattern is due to the combined effect of the diffraction of each slit + the interference of the light coming from the two slits.
The condition to observe a maximum (bright fringe), so costructive interference, in the distant screen, is:
[tex]y=\frac{m\lambda D}{d}[/tex]
where:
y is the distance of the m-th maximum from the central fringe
[tex]\lambda[/tex] is the wavelength of the light used
D is the distance of the screen from the slits
d is the separation between the slits
In this problem, we know that:
[tex]\lambda=608 nm=608\cdot 10^{-9}m[/tex] is the wavelength of light used
[tex]D=3.00 m[/tex] is the distance of the screen
[tex]y=4.84 mm = 4.84\cdot 10^{-3} m[/tex] is the distance of the first maximum (first-order bright fringe) from the central pattern, so when
m = 1
Solving for d, we find the separation of the slits:
[tex]d=\frac{m\lambda D}{y}=\frac{(1)(608\cdot 10^{-9})(3.00)}{4.84\cdot 10^{-3}}=3.77\cdot 10^{-4} m[/tex]
The first dark fringe on the screen instead is given by the formula
[tex]y'=\frac{(\frac{\lambda'}{2})D}{d}[/tex]
where
[tex]\lambda'[/tex] is the wavelength of the new light
Here we want the first dark fringe of the new light to be coincident to the first bright fringe of the previous light, so
[tex]y=4.84\cdot 10^{-3}m[/tex]
Therefore, solving for [tex]\lambda'[/tex],
[tex]\lambda'=\frac{2y'd}{D}=\frac{2(4.84\cdot 10^{-3})(3.77\cdot 10^{-4})}{3.00}=1.22\cdot 10^{-6} m = 1.22 \mu m[/tex]
