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serg [7]
3 years ago
12

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 n

ear the center of the pattern. determine the frequency of the light.
Physics
1 answer:
atroni [7]3 years ago
4 0
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
y= \frac{m \lambda D}{d}
where D=5.00 m is the distance of the screen from the slits, and 
d=0.048 mm=0.048 \cdot 10^{-3}m 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:
\lambda =  \frac{yd}{mD}= \frac{(0.065 m)(0.048 \cdot 10^{-3}m)}{(1)(5.00 m)}=  6.24 \cdot 10^{-7}m

And from the relationship between frequency and wavelength, c=\lambda f, we can find the frequency of the light:
f= \frac{c}{\lambda}= \frac{3 \cdot 10^8 m/s}{6.24 \cdot 10^{-7}m}=4.81 \cdot 10^{14}Hz
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