For a certain transverse wave, the distance between two successive crests is 1.20 m and eight additional crests pass a given poi
1 answer:
The distance between two succesive crests of a wave corresponds to its wavelength, therefore the wavelength of this wave is
The frequency of a wave is the number of crests that passes through a given point in a certain time; therefore, for this wave it is:
And now we can calculate the wave speed, which is given by the product between the wavelength and the frequency:
The frequency of a wave is the number of crests that passes through a given point in a certain time; therefore, for this wave it is:
And now we can calculate the wave speed, which is given by the product between the wavelength and the frequency:
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Answer:
Among those two medium, light would travel faster in the one with a reflection angle of (when light enters from the air.)
Explanation:
Let denote the speed of light in the first medium. Let
denote the speed of light in the air. Assume that the light entered the boundary at an angle of
to the normal and exited with an angle of
. By Snell's Law, the sine of
and
would be proportional to the speed of light in the corresponding medium. In other words:
.
When light enters a boundary at the critical angle , total internal reflection would happen. It would appear as if the angle of refraction is now
. (in this case,
.)
Substitute this value into the Snell's Law equation:
.
Rearrange to obtain an expression for the speed of light in the first medium:
.
The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.
For ,
is monotonically increasing with respect to
. In other words, for
in that range, the value of
increases as the value of
increases.
Therefore, compared to the medium in this question with , the medium with the larger critical angle
would have a larger
. such that light would travel faster in that medium.