Monochromatic light falls on two very narrow slits 0.048 mm apart. successive fringes on a screen 5.00 m away are 6.5 cm apart near the center of the pattern. determine the frequency of the light.

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

where D=5.00 m is the distance of the screen from the slits, and 
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:


And from the relationship between frequency and wavelength, , we can find the frequency of the light:

Ibelieve the answer is b. dugout

50 meters per 5 seconds