A beam of light is incident on a glass slab $\left( {\mu = 1.54} \right)$ in a direction as shown in the figure. The reflected light is analysed by a Polaroid prism. On rotating the Polaroid (take, tan 57°=1.54)
Answer
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Hint: We can measure the angle of incidence and from it, angle of polarization can be found (tan of incidence angle). On rotating the Polaroid, the intensity of the light will vary which can be found using appropriate laws: -
Brewster’s law: When $\tan i = \mu $ where $\tan i$ is angle of polarization and $\mu $ is refractive index of material, then the light passing will be completely polarized
Malus’s law: It states that intensity of light passing through the polarizer varies according to the square of cosine of angle formed between polarizer and axis (of analyser)
Complete step by step answer:
Angle of incidence is the angle made by the incident ray and the normal (of the surface).
As the normal forms 90° with the surface:
$
{33^\circ } + \angle i = {90^\circ } \\
\angle i = {90^\circ } - {33^\circ } \\
\Rightarrow \angle i = {57^\circ } \\
$
Calculating the value of tan for this angle of incidence which will be the angle of polarization:
$
\tan i = \tan {57^\circ } \\
\tan {57^\circ } = 1.54(given) \\
$
The given refractive index of glass slab $\left( {{\mu _g}} \right)$ is 1.54
$ \Rightarrow \tan i = \mu $
Now, Brewster's law states that the angle of polarization is equal to the refractive index of the medium, thus, Brewster’s law is followed here.
This shows that the reflected light is completely polarized.
From Malus’s law, the intensity of the light is:
$I = {I_0}{\cos ^2}\theta $
We can say that the intensity of the light will vary according to the square of the cosine of angle formed (between polarizer and axis of analyser whose graph is given as:
Thus, when the polaroid is rotated at any angle $\theta $, the intensity of the light varies as:
First, it gradually decreases, after reaching zero, its value increases again.
Note:Normal is a line perpendicular to the surface (makes angle of 90°) and the angle of incidence and reflection both are measured with respect to this normal only.
In the slab, not only reflection but partial refraction also takes place, but here, that is in very less amount and thus neglected.
Brewster’s law: When $\tan i = \mu $ where $\tan i$ is angle of polarization and $\mu $ is refractive index of material, then the light passing will be completely polarized
Malus’s law: It states that intensity of light passing through the polarizer varies according to the square of cosine of angle formed between polarizer and axis (of analyser)
Complete step by step answer:
Angle of incidence is the angle made by the incident ray and the normal (of the surface).
As the normal forms 90° with the surface:
$
{33^\circ } + \angle i = {90^\circ } \\
\angle i = {90^\circ } - {33^\circ } \\
\Rightarrow \angle i = {57^\circ } \\
$
Calculating the value of tan for this angle of incidence which will be the angle of polarization:
$
\tan i = \tan {57^\circ } \\
\tan {57^\circ } = 1.54(given) \\
$
The given refractive index of glass slab $\left( {{\mu _g}} \right)$ is 1.54
$ \Rightarrow \tan i = \mu $
Now, Brewster's law states that the angle of polarization is equal to the refractive index of the medium, thus, Brewster’s law is followed here.
This shows that the reflected light is completely polarized.
From Malus’s law, the intensity of the light is:
$I = {I_0}{\cos ^2}\theta $
We can say that the intensity of the light will vary according to the square of the cosine of angle formed (between polarizer and axis of analyser whose graph is given as:
Thus, when the polaroid is rotated at any angle $\theta $, the intensity of the light varies as:
First, it gradually decreases, after reaching zero, its value increases again.
Note:Normal is a line perpendicular to the surface (makes angle of 90°) and the angle of incidence and reflection both are measured with respect to this normal only.
In the slab, not only reflection but partial refraction also takes place, but here, that is in very less amount and thus neglected.
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