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stealth61 [152]
3 years ago
9

In a double-slit experiment, if the central diffraction peak contains 13 interference fringes, how many fringes are contained wi

thin each secondary diffraction peak (between m= + 1 and + 2 in Dsinθ=mλ). Assume the first diffraction minimum occurs at an interference minimum. Express your answer as an integer.
Physics
1 answer:
Ronch [10]3 years ago
6 0

Answer:

Explanation:

Width of central diffraction peak is given by the following expression

Width of central diffraction peak= 2 λ D/ d₁

where d₁ is width of slit and D is screen distance and λ is wave length.

Width of other fringes become half , that is each of  secondary diffraction fringe is equal to

λ D/ d₁

Width of central interference  peak is given by the following expression

Width of  each of  bright fringe =  λ D/ d₂

where d₂ is width of slit and D is screen distance and λ is wave length.

Now given that the central diffraction peak contains 13 interference fringes

so ( 2 λ D/ d₁)  /  λ  D/ d₂ = 13

then (  λ D/ d₁)  /  λ  D/ d₂ = 13 / 2

= 6.5

no of fringes  contained within each secondary diffraction peak = 6.5

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