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light of wavelength 485 nm passes through a single slit of width 8.32 *10^-6m. what is the single between the first (m=1) and second (m=2) interference minima?

Respuesta :

Answer:

3.35

Explanation:

Got it on Acellus

The light of wavelength 485 nm passes through a single slit. The single between the first (m=1) and second (m=2) interference minima is 3.36°.

What is diffraction?

Diffraction is the phenomenon of bending of waves through obstacles.  

Given is the wavelength λ= 485 nm, silt width d =  8.32 *10⁻⁶ m, then the angle θ will be

d sinθ =mλ

for m=1, sin θ₁ = λ/d

for m=2, sin θ₂ = 2λ/d

Substitute the values into both expressions to find the angles,

sin θ₁ = 485 x 10⁻⁹ /  8.32 *10⁻⁶

θ₁ = 3.34°

and sin θ₂ = (2 x  485 x 10⁻⁹ )/  8.32 *10⁻⁶

θ₂ = 6.7°

The angle between m =1 and m=2 will be

θ₂ -θ₁ = 6.7° - 3.34° =3.36°

Thus, angle between the first (m=1) and second (m=2) interference minima is 3.36°.

Learn more about diffraction.

https://brainly.com/question/12290582

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