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If epsilon0 and mu0 are respectively the...

If `epsilon_0` and `mu_0` are respectively the electric permittivity and the magnetic permeability of free space and `epsilon` and `mu` the corresponding quantities in a medium, the refractive index of the medium is

A

(a) `sqrt(((muepsilon)/(mu_0epsilon_0)))`

B

(b) `(muepsilon)/(mu_0epsilon_0)`

C

(c) `sqrt(((mu_0epsilon_0)/(muepsilon)))`

D

(d) `sqrt(((mumu_0)/(epsilonepsilon_0)))`

Text Solution

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The correct Answer is:
To find the refractive index of a medium in terms of the electric permittivity and magnetic permeability of free space and the corresponding quantities in the medium, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Refractive Index**: The refractive index \( n \) of a medium is defined as the ratio of the speed of light in vacuum \( c \) to the speed of light in the medium \( v \): \[ n = \frac{c}{v} \] 2. **Speed of Light in Vacuum**: The speed of light in vacuum \( c \) can be expressed in terms of the electric permittivity \( \epsilon_0 \) and magnetic permeability \( \mu_0 \) of free space: \[ c = \frac{1}{\sqrt{\mu_0 \epsilon_0}} \] 3. **Speed of Light in a Medium**: The speed of light in a medium can be expressed in terms of the electric permittivity \( \epsilon \) and magnetic permeability \( \mu \) of that medium: \[ v = \frac{1}{\sqrt{\mu \epsilon}} \] 4. **Substituting into the Refractive Index Formula**: Now, substituting the expressions for \( c \) and \( v \) into the refractive index formula: \[ n = \frac{c}{v} = \frac{\frac{1}{\sqrt{\mu_0 \epsilon_0}}}{\frac{1}{\sqrt{\mu \epsilon}}} \] 5. **Simplifying the Expression**: This simplifies to: \[ n = \frac{\sqrt{\mu \epsilon}}{\sqrt{\mu_0 \epsilon_0}} = \sqrt{\frac{\mu \epsilon}{\mu_0 \epsilon_0}} \] 6. **Final Expression for the Refractive Index**: Therefore, the refractive index \( n \) of the medium can be expressed as: \[ n = \sqrt{\frac{\mu}{\mu_0} \cdot \frac{\epsilon}{\epsilon_0}} \] ### Final Answer: The refractive index of the medium is given by: \[ n = \sqrt{\frac{\mu}{\mu_0} \cdot \frac{\epsilon}{\epsilon_0}} \]

To find the refractive index of a medium in terms of the electric permittivity and magnetic permeability of free space and the corresponding quantities in the medium, we can follow these steps: ### Step-by-Step Solution: 1. **Understanding the Refractive Index**: The refractive index \( n \) of a medium is defined as the ratio of the speed of light in vacuum \( c \) to the speed of light in the medium \( v \): \[ n = \frac{c}{v} ...
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Knowledge Check

  • If epsi_(0) and mu_(0) are respectively, the electric permittivity and the magnetic permeability of free space, epsi and mu the corresponding quantities in a medium, the refractive index of the medium is

    A
    `sqrt((mu epsi)/(mu_(0) epsi_(0)))`
    B
    `(mu epsi)/(mu_(0) epsi_(0))`
    C
    `sqrt((mu_(0) epsi_(0))/(mu epsi))`
    D
    `sqrt((mu mu_(0))/(epsi epsi_(0)))`
  • The unit of permittivity of free space epsilon_(0) is:

    A
    `"coulomb"//"newton" - "metre"`
    B
    `"newton"-metre^(2)//"coulomb"^(2)`
    C
    `"coulomb"^(2)//"newton" - "metre"^(2)`
    D
    `"coulomb"^(2)//("newton" - "metre")^(2)`
  • The unitof permittivity of free space epsilon_(0) is

    A
    coulomb `//` newton-metre
    B
    coulom`b^(2) //`newton -metre`^(2)`
    C
    newton-metr`e^(2) //`coulom`b^(2)`
    D
    coulom`b^(2) //`(newton-metre`)^(2)`
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