Question

At room temperature, if the relative permittivity of the water be 80 and the relative permeability be 0.0222, then the velocity of the light in water is ____ m/s
(A) $3 \times {10^8}$
(B) $2.25 \times {10^8}$
(C) $2.5 \times {10^8}$
(D) $3.5 \times {10^8}$

Answer 1

Hint: In this question, we need to determine the velocity of the light in the water (in meters per second) such that the relative permittivity of the water is 80 and the relative permeability is 0.0222. For this, we will write the formula for the velocity of light in air and water on the terms of permittivity and permeability and divide the results.

Complete step by step answer:
The inverse of the square root of the product of the relative permittivity and relative permeability of the water results in the speed of the light in water. Mathematically $C = \dfrac{1}{{\sqrt {{\varepsilon _r}{\mu _r}} }} - - - - (i)$.
So, the speed of light in air is given as $C' = \dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }} - -
(ii)
$ where ${\varepsilon _r}{\text{ and }}{\mu _r}$ are the absolute value of the permittivity and the permeability in air.
Divide the equation (ii) by the equation (i), we get
$
\dfrac{C}{{C'}} = \dfrac{{\left( {\dfrac{1}{{\sqrt {{\varepsilon _r}{\mu _r}} }}} \right)}}{{\left( {\dfrac{1}{{\sqrt {{\varepsilon _0}{\mu _0}} }}} \right)}} \\
\Rightarrow\dfrac{C}{{C'}}= \dfrac{{\sqrt {{\varepsilon _0}{\mu _0}} }}{{\sqrt {{\varepsilon _r}{\mu _r}} }} \\
\Rightarrow\dfrac{C}{{C'}}= \sqrt {\dfrac{{{\varepsilon _0}{\mu _0}}}{{{\varepsilon _r}{\mu _r}}}} \\
$
On cross multiplying the terms, we get
$
\dfrac{C}{{C'}} = \sqrt {\dfrac{{{\varepsilon _0}{\mu _0}}}{{{\varepsilon _r}{\mu _r}}}} \\
\Rightarrow C = C'\sqrt {\dfrac{{{\varepsilon _0}{\mu _0}}}{{{\varepsilon _r}{\mu _r}}}} \\
$
Now, substituting the values of permittivity and permeability for air as 1 and the speed of the light in the air as $3 \times {10^8}$ meters per second in the above equation, we get
$
C = C'\sqrt {\dfrac{{{\varepsilon _0}{\mu _0}}}{{{\varepsilon _r}{\mu _r}}}} \\
\Rightarrow C= 3 \times {10^8}\sqrt {\dfrac{{1 \times 1}}{{80 \times 0.0222}}} \\
\Rightarrow C= 3 \times {10^8}\sqrt {\dfrac{1}{{1.775}}} \\
\Rightarrow C= 3 \times {10^8} \times 0.75 \\
\therefore C= 2.25 \times {10^8}{\text{ m/s}} \\
$
Hence, the velocity of the light in the water (in meters per second) such that the relative permittivity of the water is 80 and the relative permeability is 0.0222 is $2.25 \times {10^8}{\text{ m/s}}$.
Hence,option B is correct.

Note:Speed of light decreases as the permittivity and permeability increases. Permittivity is the property of every material which measures the opposition offered against the formation of an electric field. The permeability plays an important role in classifying the magnetization property of a material.