Hèłp mèéëėę ...............
2 answers:
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Answer:
Frequency of oscillation, f = 4 Hz
time period, T = 0.25 s
Angular frequency,
Given:
Time taken to make one oscillation, T = 0.25 s
Solution:
Frequency, f of oscillation is given as the reciprocal of time taken for one oscillation and is given by:
f =
f =
Frequency of oscillation, f = 4 Hz
The period of oscillation can be defined as the time taken by the suspended mass for completion of one oscillation.
Therefore, time period, T = 0.25 s
Angular frequency of oscillation is given by:
1) Medium "b" has more optical density
2) Light must hit the interface between the two mediums perpendicularly
Explanation:
1)
Refraction occurs when light propagates from a medium into a second medium.
The optical density of a medium is given by its index of refraction, which is defined as:
where
c is the speed of light in a vacuum
v is the speed of light in a medium
Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.
In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.
2)
We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:
where
are the index of refraction of the two mediums
are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)
Here we want the direction of propagation of the light ray not to change: this means that it must be
(1)
However, here we have two mediums "a" and "b" with different index of refraction, so
Therefore the only angle that can satisfy eq.(1) is
So, the light must hit the surface perpendicular to the interface between the two mediums.
Learn more about refraction:
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