Answer:
38.12 mm
Explanation:
In the interference pattern created by the diffraction of light through a double slit, the position of the maxima (bright fringes) are given by
[tex]y_m = \frac{m\lambda D}{d}[/tex]
where
m is the order of the maximum
[tex]\lambda[/tex] is the wavelength of the light
D is the distance of the screen from the slits
d is the separation between the slits
On the other hand, the position of the minima (dark fringes) is given by
[tex]y_m = \frac{(m+\frac{1}{2})\lambda D}{d}[/tex]
where
(m+1) is the order of the minima
In this problem we have:
[tex]d=60.3\mu m = 60.3\cdot 10^{-6} m[/tex] is the separation between the slits
[tex]\lambda=537.0 nm = 537.0\cdot 10^{-9} m[/tex] is the wavelength of light
D = 2.14 m is the position of the screen
So, distance of the first dark fringe (m=0) from the central maximum is
[tex]y_1 = \frac{(1/2)\lambda D}{d}=\frac{(537.0\cdot 10^{-9})(2.14)}{2(60.3\cdot 10^{-6})}=9.53\cdot 10^{-3} m[/tex]
On the other side, the distance of the second dark fringe (m=1) from the central maximum is
[tex]y_2= \frac{(3/2)\lambda D}{d}=\frac{3(537.0\cdot 10^{-9})(2.14)}{2(60.3\cdot 10^{-6})}=28.59\cdot 10^{-3} m[/tex]
Therefore, the distance between the two dark fringes is:
[tex]d' = y_1 + y_2 = 9.53\cdot 10^{-3} + 28.59\cdot 10^{-3} = 38.12 \cdot 10^{-3} m = 38.12 mm[/tex]
