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Tema [17]
4 years ago
10

A ray of light is moving from a material having a high index of refraction into a material with a lower index of refraction.(a)

Is the ray bent toward the normal or away from it?toward the normalaway from the normal (b) If the wavelength is 600 nm in the material with the high index of refraction, is it greater, smaller, or the same in the material with the lower index of refraction?greatersmaller the same(c) How does the frequency change as the light moves between the two materials?Does it increase, decrease, or remain the same?increasedecreaseremain the same
Physics
1 answer:
Sveta_85 [38]4 years ago
5 0

(a) Away from the normal

The direction of bending of the ray of light can be found by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n1, n2 are the index of refraction of the first and second medium, respectively

\theta_1, \theta_2 are the angle of the incident and refracted ray with respect to the normal to the surface, respectively

In this problem, the ray of light moves from a material with high index of refraction to a material with lower index, so:

n_1 > n_2

Re-arranging Snell's law we find

sin \theta_2 = \frac{n_1}{n_2} sin \theta

since

\frac{n_1}{n_2}>1

we also have

sin \theta_2 > sin \theta_1\\\theta_2 > \theta_1

so the ray of light bends away from the normal.

(b) The wavelength is greater in the second material (the one with lower index of refraction)

The wavelength of the light in a medium is given by

\lambda=\frac{\lambda_0}{n}

where

\lambda_0 is the wavelength of the light in a vacuum

n is the refractive index

We can rewrite the equation as

\lambda_0 = \lambda_1 n_1 = \lambda_2 n_2

And isolating \lambda_2 from the second equation

\lambda_2 = \frac{n_1}{n_2} \lambda_1

where

\lambda_1 = 600 nm\\\frac{n_1}{n_2}>1

So, we have that the wavelength in the second medium (the one with lower index of refraction) is longer than the wavelength in the first medium.

(c) The frequency is the same

While wavelength and speed of a light wave depend on the medium in which the wave is travelling through, the frequency does not depend on that, so it is exactly the same in the two mediums.

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