
The speed of light is \[299,792,458{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\]
This question has multiple correct options
A. with respect to the earth
B. with respect to the sun
C. with respect to a train moving on the earth
D. with respect to a spaceship going in outer space
Answer
580.8k+ views
Hint: We know that the expression of speed of light is inversely proportional to the square root of the product of magnetic permeability and permittivity of free space. We can calculate the speed of light using this expression to get the exact answer.
Complete step by step answer:We know that the speed of light is \[299,792,458{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\] , and we can write its approximate value equal to \[3 \times {10^8}{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\]. We also know that light's speed is independent of the respect with which it is being measured. In other words, we can say that speed of light is independent of reference point and respect of being measured.
Therefore, we can say that speed of light with respect to the earth, sun, a train moving on the earth, and a spaceship going in outer space will be given by \[299,792,458{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\]. Hence all the given options (A), (B), (C) and (D) are correct.
Additional information: Magnetic permeability is the resistance provided by the material for the magnetic field to take place in it. The permittivity of free space is the ability of vacuum to allow an electric field to take place.
Note:We can derive the expression for the speed of light by using the concept of Maxwell's equations. The final expression after calculations comes out to be as given below:
\[c = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}\]
Here c is the speed of light, \[{\mu _0}\] is magnetic permeability, and \[{\varepsilon _0}\] is permittivity in free space or in vacuum. We can remember the approximate value of the speed of light that can be used in numerical problems.
Complete step by step answer:We know that the speed of light is \[299,792,458{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\] , and we can write its approximate value equal to \[3 \times {10^8}{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\]. We also know that light's speed is independent of the respect with which it is being measured. In other words, we can say that speed of light is independent of reference point and respect of being measured.
Therefore, we can say that speed of light with respect to the earth, sun, a train moving on the earth, and a spaceship going in outer space will be given by \[299,792,458{\rm{ }}{{\rm{m}} {\left/
{\vphantom {{\rm{m}} {\rm{s}}}} \right.
} {\rm{s}}}\]. Hence all the given options (A), (B), (C) and (D) are correct.
Additional information: Magnetic permeability is the resistance provided by the material for the magnetic field to take place in it. The permittivity of free space is the ability of vacuum to allow an electric field to take place.
Note:We can derive the expression for the speed of light by using the concept of Maxwell's equations. The final expression after calculations comes out to be as given below:
\[c = \dfrac{1}{{\sqrt {{\mu _0}{\varepsilon _0}} }}\]
Here c is the speed of light, \[{\mu _0}\] is magnetic permeability, and \[{\varepsilon _0}\] is permittivity in free space or in vacuum. We can remember the approximate value of the speed of light that can be used in numerical problems.
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