Answer :
In a double-slit interference experiment, the distance y of the maximum of order m from the center of the observed interference pattern on the screen is
[tex]y= \frac{m \lambda D}{d} [/tex]
where D=5.00 m is the distance of the screen from the slits, and
[tex]d=0.048 mm=0.048 \cdot 10^{-3}m[/tex] is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
[tex]\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m[/tex]
And from the relationship between frequency and wavelength, [tex]c=\lambda f[/tex], we can find the frequency of the light:
[tex]f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz [/tex]
[tex]y= \frac{m \lambda D}{d} [/tex]
where D=5.00 m is the distance of the screen from the slits, and
[tex]d=0.048 mm=0.048 \cdot 10^{-3}m[/tex] is the distance between the two slits.
The fringes on the screen are 6.5 cm=0.065 m apart from each other, this means that the first maximum (m=1) is located at y=0.065 m from the center of the pattern.
Therefore, from the previous formula we can find the wavelength of the light:
[tex]\lambda = \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}= 6.24 \cdot 10^{-7}m[/tex]
And from the relationship between frequency and wavelength, [tex]c=\lambda f[/tex], we can find the frequency of the light:
[tex]f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz [/tex]