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# The critical angle of medium for specific wavelength, if the medium has relative permittivity $3$ and relative permeability $\dfrac{4}{3}$ for this wavelength, will be:A) ${45^ \circ }$B) ${30^ \circ }$C) ${15^ \circ }$D) ${60^ \circ }$

Last updated date: 03rd Aug 2024
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Hint: Use the formula of the refractive index of the medium and substitute the formula of velocity of light in air and medium. Substitute the angles, and the obtained refractive index of the medium in the snell’s law to know the critical angle of the medium.

Useful formula:
(1) The relative permittivity is given by
${ \in _r} = \dfrac{{{ \in _{}}}}{{{ \in _0}}}$
Where ${ \in _0}$ is the permittivity of free space and $\in$ is the permittivity of the medium.

(2) The relative permeability of the medium is given by
${\mu _r} = \dfrac{{{\mu _{}}}}{{{\mu _0}}}$
Where $\mu$ is the permeability of the medium and ${\mu _0}$ is the permeability of the free space.

(3) The refractive index of the medium is given by
${\mu _2} = \dfrac{c}{v}$
Where $c$ is the velocity of the light in vacuum and $v$ is the velocity of the light in medium.

(4) The snell’s law states that
${\mu _2}\sin {\theta _i} = {\mu _1}\sin {\theta _r}$
Where ${\mu _1}$ is the refractive index of free space and ${\mu _2}$ is the refractive index of the medium.

Complete step by step solution:
It is given that the
Relative permittivity of the medium, ${ \in _r} = 3$
The relative permeability of the medium, $\mu = \dfrac{4}{3}$
By taking the formula (3),
${\mu _2} = \dfrac{c}{v}$
Substituting the values of $c = \dfrac{1}{{\sqrt {{v_o}{ \in _0}} }}$ and the $v = \dfrac{1}{{\sqrt {\mu { \in _r}} }}$ in the above formula,

${\mu _2} = \dfrac{{\dfrac{1}{{\sqrt {{v_o}{ \in _0}} }}}}{{\dfrac{1}{{\sqrt {\mu { \in _r}} }}}}$
By simplifying the above equation, and also using the formula (1) and (2) in it, we get
${\mu _2} = \sqrt {{\mu _r}{ \in _r}}$
${\mu _2} = \sqrt {3 \times \dfrac{4}{3}}$
${\mu _2} = 2$
Using the formula (4),
${\mu _2}\sin {\theta _i} = {\mu _1}\sin {\theta _r}$
The critical angle ${\theta _r} = {90^ \circ }$, so
${\mu _2}\sin {\theta _i} = {\mu _1}\sin {90^ \circ }$
${\mu _2}\sin {\theta _i} = 2 \times \dfrac{1}{2}$
Substituting the value of the angles and the refractive index of the medium
$2\sin {\theta _i} = 1$
$\sin {\theta _i} = \dfrac{1}{2}$
Hence the value of the critical angle of the medium is ${30^ \circ }$.

Thus the option (B) is correct.

Note: The snell’s law has the relation, in which the ratio of the sine of the angles of incidence and the refraction is equal to the ratio of the refractive indexes. It is mainly used in fiber optics. Always remember that the critical angle of the free space is ${90^ \circ }$ .