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Two polaroid are kept crossed to each other. If one of them is rotated at an angle \[60{}^\circ ,\] the percentage of incident light now transmitted through the system is.

Last updated date: 17th Jun 2024
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Hint: When the two polaroid is kept across to each other one of them have rotated angle of $60{}^\circ $, So, we have the intensity of polarized light as ${{I}_{0}}$ so the intensity of the first polaroid is $\dfrac{{{I}_{0}}}{2}$. Using the given information the equation. $I={{I}_{0}}{{\cos }^{2}}\theta $. Which can be written as ${{I}_{0}}=\dfrac{{{I}_{0}}}{2}{{\cos }^{2}}\theta $ and it used for finding the answer.

Complete step by step solution:When the two polaroid are kept crossed to each other. If one of them is rotated at an angle of \[60{}^\circ \]
The intensity of un polarized light be ${{I}_{0}},$ So, the intensity of the first polaroid is $\dfrac{{{I}_{0}}}{2}$
On rotating through $60{}^\circ .$ The intensity of light from sound polaroid.
As we have, $I={{I}_{0}}{{\cos }^{2}}\theta $
$\therefore I=\left( \dfrac{{{I}_{0}}}{2} \right){{\left( \cos 60{}^\circ \right)}^{2}}$
$=\dfrac{{{I}_{0}}}{2}\times \dfrac{1}{4}$
$\therefore I=0.125{{I}_{0}}$
Percentage of incident light transmitted through the system $=0.125\times 100=12.5%$
The required percentage is $12.5%$

Additional Information:
Polarization is a property that is applying to the transverse waves which specifies that orientation of oscillations. The direction of oscillation is perpendicular in a transverse wave. For an example for a polarized transverse wave that is vibration traveling along the string. (or a guitar string). The vibrations can be in horizontal or vertical direction. Depends upon how string is Plock. Some of the optical measurement steps are based on polarization. Polarization is crucial or it must be taken into account and it is controlled. Such examples are numbers for mention.

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