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Katyanochek1 [597]
4 years ago
12

Light from a laser with wavelength 400 nm passed through two small openings and produces an interference pattern on a screen 1 m

away. If the first point of destructive interference occurs at an angle of 0.1 radians, what is the distance between the two openings?
Physics
1 answer:
alexira [117]4 years ago
8 0

To solve this problem we will apply the concepts related to destructive interference from double-slit experiments. For this purpose we will define the path difference as,

\text{Path difference}= dsin\theta = (2n-1)\frac{\lambda}{2}

Here,

\lambda = Wavelength

\theta = Angle when occurs the interference point of destructive interference

Our values are given as,

\text{Wavelength} = \lambda = 400nm = 4*10^{-7}m

\text{Distance of Screen} = D = 1m

Using the previous expression we have,

d \times \theta = \frac{\lambda}{2}

d \times (0.1) = \frac{4*10^{-7}}{2}

d = 2*10^{-6} m

d = 2\mu md = 2\mu m

Therefore the distance between the two openings is 2\mu m

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