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Coherent light of frequency f travels in air and is incident on two narrow slits. The interference pattern is observed on a distant screen that is directly opposite the slits. The frequency of light f can be varied. For f = 5.60 × 1012 Hz there is an interference maximum for e = 60.0°. The next higher frequency for which there is an interference maximum at this Part A What is the separation d between the two slits? angle is 7.47 x 1012 Hz Express your answer with the appropriate units. HÀ Value d = _______.

Respuesta :

Answer:

A)        d = 6.1857 10⁻⁵ m

Explanation:

In the double-slit experiment the constructive interference is given by the expression

          d sin θ = m λ

          d =[tex]\frac{m \lambda }{sin \theta}[/tex]

let's use the relationship between speed, wavelength and frequency

          c = λ f

          λ = c / f

          λ = 3 10⁸ / 5.60 10¹²

          λ = 5.357 10⁻⁵ m

I do not know India in order of interference, we will assume that  m= 1

          d = 1  5,357 10⁻⁵ / sin 60

          d = 6.1857 10⁻⁵ m

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