Light travels through a liquid at 2.25e8 m/s. What is the index of refraction of the liquid?
2 answers:
Answer:
The index of refraction of the liquid is n = 1.33 equivalent to that of water
Explanation:
Solution:-
- The index of refraction of light in a medium ( n ) determines the degree of "bending" of light in that medium.
- The index of refraction is material property and proportional to density of the material.
- The denser the material the slower the light will move through associated with considerable diffraction angles.
- The lighter the material the faster the light pass through the material without being diffracted as much.
- So, in the other words index of refraction can be expressed as how fast or slow light passes through a medium.
- The reference of comparison of how fast or slow the light is the value of c = 3.0*10^8 m/s i.e speed of light in vacuum or also assumed to be the case for air.
- so we can mathematically express the index of refraction as a ratio of light speed in the material specified and speed of light.
- The light passes through a liquid with speed v = 2.25*10^8 m/s :
- The index of refraction of the liquid is n = 1.33 equivalent to that of water.
Answer:
Refractive index = 1.33
Explanation:
Refractive index is of a material explains how light passes through a medium. It is the ratio of speed of light in vacuum to the speed of light in the medium.
We know the speed of light in water is and there is a change iñ speed of light in every medium it passes (except when it passes through vacuum). The change in speed of light is known to be =
Given speed of light = 2.25*10^8 m/s
Formula for refractive index:
Where,
change in speed of light, c=
speed of light, v = 2.25*10^8 m/s
= 1.328
= 1.33
The liquid is water, and the index of refraction of water is 1.33
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Complete question:
Point charges q1=- 4.10nC and q2=+ 4.10nC are separated by a distance of 3.60mm , forming an electric dipole. The charges are in a uniform electric field whose direction makes an angle 36.8 ∘ with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude 7.30×10−9 N⋅m ? Express your answer in newtons per coulomb to three significant figures.
Answer:
The magnitude of this field is 826 N/C
Explanation:
Given;
The torque exerted on the dipole, T = 7.3 x 10⁻⁹ N.m
PEsinθ = T
where;
E is the magnitude of the electric field
P is the dipole moment
First, we determine the magnitude dipole moment;
Magnitude of dipole moment = q*r
P = 4.1 x 10⁻⁹ x 3.6 x 10⁻³ = 1.476 x 10⁻¹¹ C.m
Finally, we determine the magnitude of this field;
E = 826 N/C (in three significant figures)
Therefore, the magnitude of this field is 826 N/C