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pentagon [3]
2 years ago
13

In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the

first light source has a wavelength of 629 nm. Two different interference patterns are observed. If the 7th order bright fringe from the first light source coincides with the 8th order bright fringe from the second light source, what is the wavelength of the light coming from the second monochromatic light source
Physics
1 answer:
GaryK [48]2 years ago
7 0

We will determine the wavelength through the relationship given by the distance between slits, this relationship is given under the function

y = \frac{m\lambda}{d}

Here,

m = Number of order bright fringe

\lambda = Wavelength

d = Distance between slits

Both distance are the same, then

y_1 = y_2

\frac{m_1\lambda_1 r}{d} = \frac{m_2\lambda_2 r}{d}

\frac{m_1\lambda_1}{m_2\lambda_2} =1

 Rearranging to find the second wavelength

m_1 \lambda_1 = m_2 \lambda _2

\lambda_2 = \frac{m_1\lambda_1}{m_2}

\lambda_2 = \frac{7(629)}{8}

\lambda_2 = 550.3nm

Therefore the wavelength of the light coming from the second monochromatic light source is 550.3nm

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A light ray moving from oil (n=1.38) into an unknown material. If the light is incident at 35 and refracts at 23, what is the in
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The index of refraction of the unknown material in which a ray of light is incident at 35° and refracted at 23° is 2.03

<h3>Snell's law</h3>

index of refraction (n) = Sine i / Sine r

n = Sine i / Sine r

Where

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  • r is the angle of refraction
<h3>How to determine the refractive index </h3>

From the question given above, the following data were obtained:

  • Index of refraction of oil (nₒ) = 1.38
  • Angle of incidence (i) = 35°
  • Angle of refraction (r) = 23°
  • Index of refraction of unknown material (nᵣ) =?

nₒSine i = nᵣSine r

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Divide both side by Sine 23

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Thus, the index of refraction of the unknown material is 2.03

Learn more about Snell's law:

brainly.com/question/25758484

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