
Choose the correct alternative give.
If ${\varepsilon _ \circ }$ and ${\mu _ \circ }$ are respectively the electric permittivity and the magnetic permeability of free space and $\varepsilon $and $\mu $ be the respective quantities in a medium, the refractive index of the medium is:
A) $\sqrt {\dfrac{{\mu \varepsilon }}{{{\mu _ \circ }{\varepsilon _ \circ }}}} $
B) $\dfrac{{\mu \varepsilon }}{{{\mu _ \circ }{\varepsilon _ \circ }}}$
C) $\sqrt[{}]{{\dfrac{{{\mu _ \circ }{\varepsilon _ \circ }}}{{\mu \varepsilon }}}}$
D) $\dfrac{{\mu {\mu _ \circ }}}{{\varepsilon {\varepsilon _ \circ }}}$
Answer
216.6k+ views
Hint: The ratio of speed of light in free space to the speed of light in medium is called refractive index of the medium.
Complete step by step solution:
The refractive index of a medium is given by, RI=$\dfrac{c}{v}$ --------- (equation: 1)
Here, RI = refractive index of the medium, c= speed of light in free space, v= speed of light in the medium.
Also speed of any electromagnetic wave in a medium is given by the formula,
v = $\dfrac{1}{{\sqrt {\varepsilon \mu } }}$ ---- (here, v= speed of em wave in the medium,$\varepsilon _{}^{}$= electric permittivity of the medium, $\mu $= magnetic permeability of the medium.) ------ (equation: 2)
Now, it is given that electric permittivity for free space is ${\varepsilon _ \circ }$ and magnetic permeability for free space is ${\mu _ \circ }$ .
And the electric permittivity and magnetic permeability in the medium are $\varepsilon $and $\mu $ respectively.
Let c be the speed of light in free space and v be the speed of light in medium, then,
$c = \dfrac{1}{{\sqrt[{}]{{{\varepsilon _ \circ }{\mu _ \circ }}}}}$ ----------- (from equation: 2)----- (equation:3)
Also, $v = \dfrac{1}{{\sqrt {\varepsilon \mu } }}$ ----------- (from equation: 2) ------ (equation: 4)
Thus, refractive index of the medium is given by,
RI = $\dfrac{c}{v}$ ----- (from equation: 1)
RI = $\dfrac{1}{{\sqrt {{\varepsilon _ \circ }{\mu _ \circ }} }} \times \sqrt {\varepsilon \mu } $ --------- (from equation: 3 and equation: 4)
Therefore, RI = $\dfrac{{\sqrt {\varepsilon \mu } }}{{\sqrt {{\varepsilon _ \circ }{\mu _ \circ }} }}$
Hence, option A is correct.
Note: Here, the electric permittivity and magnetic permeability of the mediums is not with respect to that in free space.
If relative magnetic permeability and electric permittivity were given then the refractive index would have been; \[RI = \sqrt {{\varepsilon _r}{\mu _r}} \]. Here, ${\varepsilon _r}$ is the relative electric permittivity and ${\mu _r}$ is the relative magnetic permeability.
Complete step by step solution:
The refractive index of a medium is given by, RI=$\dfrac{c}{v}$ --------- (equation: 1)
Here, RI = refractive index of the medium, c= speed of light in free space, v= speed of light in the medium.
Also speed of any electromagnetic wave in a medium is given by the formula,
v = $\dfrac{1}{{\sqrt {\varepsilon \mu } }}$ ---- (here, v= speed of em wave in the medium,$\varepsilon _{}^{}$= electric permittivity of the medium, $\mu $= magnetic permeability of the medium.) ------ (equation: 2)
Now, it is given that electric permittivity for free space is ${\varepsilon _ \circ }$ and magnetic permeability for free space is ${\mu _ \circ }$ .
And the electric permittivity and magnetic permeability in the medium are $\varepsilon $and $\mu $ respectively.
Let c be the speed of light in free space and v be the speed of light in medium, then,
$c = \dfrac{1}{{\sqrt[{}]{{{\varepsilon _ \circ }{\mu _ \circ }}}}}$ ----------- (from equation: 2)----- (equation:3)
Also, $v = \dfrac{1}{{\sqrt {\varepsilon \mu } }}$ ----------- (from equation: 2) ------ (equation: 4)
Thus, refractive index of the medium is given by,
RI = $\dfrac{c}{v}$ ----- (from equation: 1)
RI = $\dfrac{1}{{\sqrt {{\varepsilon _ \circ }{\mu _ \circ }} }} \times \sqrt {\varepsilon \mu } $ --------- (from equation: 3 and equation: 4)
Therefore, RI = $\dfrac{{\sqrt {\varepsilon \mu } }}{{\sqrt {{\varepsilon _ \circ }{\mu _ \circ }} }}$
Hence, option A is correct.
Note: Here, the electric permittivity and magnetic permeability of the mediums is not with respect to that in free space.
If relative magnetic permeability and electric permittivity were given then the refractive index would have been; \[RI = \sqrt {{\varepsilon _r}{\mu _r}} \]. Here, ${\varepsilon _r}$ is the relative electric permittivity and ${\mu _r}$ is the relative magnetic permeability.
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