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g How many diffraction maxima are contained in a region of the Fraunhofer single-slit pattern, subtending an angle of 2.12°, for a slit width of 0.110 mm, using light of wavelength 582 nm?

Respuesta :

Manetho

Answer:

2

Explanation:

We know that in the Fraunhofer single-slit pattern,

maxima is given by

[tex]a\text{sin}\theta=\frac{2N+1}{2}\lambda[/tex]

Given values

θ=2.12°

slit width a= 0.110 mm.

wavelength λ= 582 nm

Now plugging values to calculate N we get

[tex]0.110\times10^{-3}\text{sin}2.12=(\frac{2N+1}{2})582\times10^{-9}[/tex]

Solving the above equation we get

we N= 2.313≅ 2

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