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Light of wavelength 500 nm illuminates parallel slits and produces an interference pattern on a screen that is 1 m from the slits. In terms of the initial intensity I0, the light's intensity in the interference pattern at a point for which the path difference is 300 nm is

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Answer:

this point we have a destructive inference  m=1

Explanation:

This is an interference exercise, where the light has two possible states one of constrictive interference

            d sin θ = m λ

and another of destructive interference

           d sin θ = (m + ½) λ

the term on the left is the difference in trajectory that tells us that it is worth

            d sin θ = 300 nm = 300 10⁻⁹ m

We can find what type of interference we have by finding m, which must be an integer

             

constructive interference

           m = d sin θ /λ

let's calculate

          m = 300 10⁻⁹ / 500 10⁻⁹

          m = 0.6

this value is not possible because m must be an integer

destructive interference

          m + ½ = d sin θ /λ  

          m = d sin θ /λ  + ½

          m = 0.6 + 0.5

         m = 1.1

that can be approximated to m = 1

this value if possible, so at this point we have a destructive inference

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