Monochromatic light falls on two very narrow slits 0.046 mm apart. Successive fringes on a screen 6.20 m away are 8.9 cm apart near the center of the pattern. Part A Determine the wavelength of the light.

λ = 6.602 x 10^(-7) m

Explanation:

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 given as ;

y = mλD/d

Where;

D is the distance of the screen from the slits = 6.2 m

d is the distance between the two slits = 0.046 mm = 0.046 x 10^(-3) m

The fringes on the screen are 8.9 cm = 0.089 m apart from each other, this means that the first maximum (m=1) is located at y = 0.089 m from the center of the pattern.

Therefore, from the previous formula we can find the wavelength of the light:

y = mλD/d

So, λ = dy/mD

Thus,

λ = (0.046 x 10^(-3) x 0.089)/(1 x 6.2)

λ = 6.602 x 10^(-7) m

this would be about 43%