
Unpolarised light of intensity $I$ passes through an ideal polarizer A. Another identical polarizer B is placed behind A. The intensity of light beyond B is found to be $\dfrac{1}{2}$. Now another identical polarizer C is placed between A and B. The intensity beyond B is now found to be $\dfrac{1}{8}$. The angle between polarizer A and C is:
A) $45^\circ $
B) $60^\circ $
C) $0^\circ $
D) $30^\circ $
Answer
134.4k+ views
Hint: Whenever a light passes through a polarizer, the intensity of the light will be halved. And if there is no change in the intensity between two polarizers, they are placed parallel to each other.
Complete step by step answer:
Let’s discuss the first case, that is an unpolarised light passes through a polarizer A. we know, whenever a light passes through a polarizer, the intensity of the light will be halved.
That is $\dfrac{I}{2}$
As described in the question Light passes through two polarizers A and B, and the intensity is reduced to $\dfrac{I}{2}$
The light coming from the polarizer A is having the intensity equal to $\dfrac{I}{2}$, the same light passing through the polarizer B and the intensity remains the same.
By applying Malus formula, $I = {I_0}{\cos ^2}\theta $
Where, $I$ is the final intensity, ${I_0}$ is the intensity of light which coming from the first polarizer, $\theta $ is the angle between the two polarizers (here,\[{\theta _{AB}}\])
\[ \Rightarrow \dfrac{I}{2} = \dfrac{I}{2}{\cos ^2}{\theta _{AB}}\]
$ \Rightarrow {\cos ^2}{\theta _{AB}} = 1$
$ \Rightarrow {\theta _{AB}} = 0^\circ .$
Which means the polarizers A&B are parallel to each other.
Now another polarizer C is placed between the polarizers A&B, then the resultant becomes $\dfrac{1}{8}$.
That is the intensities coming from A=$\dfrac{I}{2}$, C=${I_C}$& B=$\dfrac{1}{8}$
By applying Malus formula between polarizers A&C
………………………………………… (Eqn. P)
By applying Malus formula between Polarizers C&B
\[ \Rightarrow \dfrac{I}{8} = {I_C}{\cos ^2}{\theta _{CB}}\]………………………………………… (Eqn. Q)
By analyzing these data, we understand that polarizer C is making angle between polarizers A&B
As A&B are parallel, ${\theta _{AC}} = {\theta _{BC}} = \theta $
Eqn. P in Eqn. Q, we get,
$ \Rightarrow \dfrac{I}{8} = \dfrac{I}{2}{\cos ^2}\theta {\cos ^2}\theta $
$ \Rightarrow \dfrac{1}{4} = {\cos ^4}\theta $
$ \Rightarrow \dfrac{1}{2} = {\cos ^2}\theta $
$ \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow \theta = 45^\circ $
We get, final answer is option (A)
Note: An unpolarized light is a light wave that is vibrating in more than one plane is referred to as unpolarized light. In unpolarised light, the vibrations are symmetric about the direction of propagation. For an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation.
Complete step by step answer:
Let’s discuss the first case, that is an unpolarised light passes through a polarizer A. we know, whenever a light passes through a polarizer, the intensity of the light will be halved.
That is $\dfrac{I}{2}$
As described in the question Light passes through two polarizers A and B, and the intensity is reduced to $\dfrac{I}{2}$
The light coming from the polarizer A is having the intensity equal to $\dfrac{I}{2}$, the same light passing through the polarizer B and the intensity remains the same.
By applying Malus formula, $I = {I_0}{\cos ^2}\theta $
Where, $I$ is the final intensity, ${I_0}$ is the intensity of light which coming from the first polarizer, $\theta $ is the angle between the two polarizers (here,\[{\theta _{AB}}\])
\[ \Rightarrow \dfrac{I}{2} = \dfrac{I}{2}{\cos ^2}{\theta _{AB}}\]
$ \Rightarrow {\cos ^2}{\theta _{AB}} = 1$
$ \Rightarrow {\theta _{AB}} = 0^\circ .$
Which means the polarizers A&B are parallel to each other.
Now another polarizer C is placed between the polarizers A&B, then the resultant becomes $\dfrac{1}{8}$.
That is the intensities coming from A=$\dfrac{I}{2}$, C=${I_C}$& B=$\dfrac{1}{8}$
By applying Malus formula between polarizers A&C
………………………………………… (Eqn. P)
By applying Malus formula between Polarizers C&B
\[ \Rightarrow \dfrac{I}{8} = {I_C}{\cos ^2}{\theta _{CB}}\]………………………………………… (Eqn. Q)
By analyzing these data, we understand that polarizer C is making angle between polarizers A&B
As A&B are parallel, ${\theta _{AC}} = {\theta _{BC}} = \theta $
Eqn. P in Eqn. Q, we get,
$ \Rightarrow \dfrac{I}{8} = \dfrac{I}{2}{\cos ^2}\theta {\cos ^2}\theta $
$ \Rightarrow \dfrac{1}{4} = {\cos ^4}\theta $
$ \Rightarrow \dfrac{1}{2} = {\cos ^2}\theta $
$ \Rightarrow \cos \theta = \dfrac{1}{{\sqrt 2 }}$
$ \Rightarrow \theta = 45^\circ $
We get, final answer is option (A)
Note: An unpolarized light is a light wave that is vibrating in more than one plane is referred to as unpolarized light. In unpolarised light, the vibrations are symmetric about the direction of propagation. For an unpolarised wave the displacement will be randomly changing with time though it will always be perpendicular to the direction of propagation.
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