
A polarising angle of glass is ${57^0}$. A ray of light which is incident at this angle will have an angle of refraction as :
A) ${33^0}$
B) ${38^0}$
C) ${25^0}$
D) ${43^0}$
Answer
584.7k+ views
Hint:We will equate Brewster law and Snell’s law and find the relation between angle of refraction and polarisation or incidence and using that relation we can find out angle of refraction:
$r = {90^0} - {i_p}$
Complete step by step solution :
Polarising angle is the angle at which an unpolarized light or other electromagnetic radiation must be incident upon a non-metallic surface for the reflected radiation to acquire maximum plane polarization.
Polarizing angle is also Brewster angle.
Brewster Law:
Tangent of polarising angle is equal to the refractive index of the reflecting medium with respect to surroundings. $\mu = \tan \theta $or \[\] $\dfrac{{\sin {i_p}}}{{\cos {i_p}}} = \mu $ ,where $\mu $ is the refractive index of the medium and $\theta $ is the polarising angle, ${i_p}$ is the angle of polarization.
Snell’s Law:
Product of the index of refraction by the sine of the angle of incidence constant for any ray of light striking the separating surface of two mediums.
$\dfrac{{\sin {i_p}}}{{\sin r}} = \mu $ ($r$ is the angle of refraction )
Snell’s Law : $\dfrac{{\sin {i_p}}}{{\sin r}} = \mu $ .............. 1
Brewster Law: $\dfrac{{\sin {i_p}}}{{\cos {i_p}}} = \mu $ ............2
Equation the above two equations we have :
$
\Rightarrow \dfrac{{\sin {i_p}}}{{\sin r}} = \dfrac{{\sin {i_p}}}{{\cos {i_p}}} \\
\Rightarrow \sin r = \cos {i_p} \\
\Rightarrow r = ({90^0} - {i_p}) \\
\\
$
In question we have angle of incidence which is :${57^0}$
Angle of refraction $r$ is given by :
${90^0} - {i_p}$
Therefore,
$
\Rightarrow r = {90^0} - {57^0} \\
\Rightarrow r = {33^0} \\
$
Note:
From the figure drawn in the solution, we can conclude that angle of incidence and angle of refraction are both equal, which is also the property of plane mirrors (used by us in our day to day life, in dressing rooms, in our hand bags etc ).
$i = r$
Both these angles becomes when two conditions got fulfilled:
- Both the mediums have the same refractive index.
- If the incidence ray falls normally to the surface of the second medium.
$r = {90^0} - {i_p}$
Complete step by step solution :
Polarising angle is the angle at which an unpolarized light or other electromagnetic radiation must be incident upon a non-metallic surface for the reflected radiation to acquire maximum plane polarization.
Polarizing angle is also Brewster angle.
Brewster Law:
Tangent of polarising angle is equal to the refractive index of the reflecting medium with respect to surroundings. $\mu = \tan \theta $or \[\] $\dfrac{{\sin {i_p}}}{{\cos {i_p}}} = \mu $ ,where $\mu $ is the refractive index of the medium and $\theta $ is the polarising angle, ${i_p}$ is the angle of polarization.
Snell’s Law:
Product of the index of refraction by the sine of the angle of incidence constant for any ray of light striking the separating surface of two mediums.
$\dfrac{{\sin {i_p}}}{{\sin r}} = \mu $ ($r$ is the angle of refraction )
Snell’s Law : $\dfrac{{\sin {i_p}}}{{\sin r}} = \mu $ .............. 1
Brewster Law: $\dfrac{{\sin {i_p}}}{{\cos {i_p}}} = \mu $ ............2
Equation the above two equations we have :
$
\Rightarrow \dfrac{{\sin {i_p}}}{{\sin r}} = \dfrac{{\sin {i_p}}}{{\cos {i_p}}} \\
\Rightarrow \sin r = \cos {i_p} \\
\Rightarrow r = ({90^0} - {i_p}) \\
\\
$
In question we have angle of incidence which is :${57^0}$
Angle of refraction $r$ is given by :
${90^0} - {i_p}$
Therefore,
$
\Rightarrow r = {90^0} - {57^0} \\
\Rightarrow r = {33^0} \\
$
Note:
From the figure drawn in the solution, we can conclude that angle of incidence and angle of refraction are both equal, which is also the property of plane mirrors (used by us in our day to day life, in dressing rooms, in our hand bags etc ).
$i = r$
Both these angles becomes when two conditions got fulfilled:
- Both the mediums have the same refractive index.
- If the incidence ray falls normally to the surface of the second medium.
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