Answer :
The width of central interference maximum is 16000×10⁻⁹m if wavelength 400nm passes through two narrow slits.
The width of the central maximum is two times as a significant part of the other maxima. For the m t h request the essential diffraction grinding condition is m λ = d sin θ m where d is the distance between adjoining cuts and the way contrast between neighboring cuts is m λ. Cut is enlightened by the illumination of wavelength 400 A ˚.
We know that width of central maximum is represented by β ,which is equal to=2λD/d
where λ is defined as the wavelength of the light wave,
D is defined as the distance of screen from central maximum
and d is defined between the slits.
So, we have λ=400nm=400×10⁻⁹m, D=4m,d=0.200mm=0.200 ˣ 10⁻³m
Now,on putting the values in above formula, we get
=>β =2×400×10⁻⁹m×4m/0.200 ˣ 10⁻³m
=>β=3200×10⁻⁹m² / 0.200 ˣ 10⁻³m
=>β=3200000×10⁻⁹m²/200 ˣ 10⁻³m
=>β=16000×10⁻⁹m
Hence, width of central maximum is 16000×10⁻⁹m.
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