A plane polarized light having intensity ${I}_{0}$ has vibration parallel to the polarizer axis. If the plane of the polarizer is rotated by $15°$. The intensity of light transmitted by polarizer is:
$A. { I }_{ 0 }\left( \dfrac { 2+\sqrt { 3 } }{ 4 } \right)$
$B. { I }_{ 0 }\left( \cfrac { 2-\sqrt { 3 } }{ 4 } \right)$
$C. { \dfrac { \sqrt { 3 } }{ 2 } { I }_{ 0 } }$
$D. { \dfrac { 3 }{ 4 } { I }_{ 0 } }$

Hint: To solve this question, use the formula for Malus’s law. It states that when completely plane polarized light incidents on an analyzer, the intensity of the light transmitted by the polarizer is directly proportional to the square of the cosine of angle between transmission axes of polarizer and analyzer. Substitute the value in the equation and find the factor by which intensity decreases.

Formula used:
$I ={I}_{0} {\cos}^{2}{\theta}$

Complete answer:
Given: $\theta$ = 15°
Malus’s law is given by,
$I ={I}_{0} {\cos}^{2}{\theta}$ …(1)
Where, I is the intensity after polarization
            ${I}_{0}$ is the initial intensity
Substituting the value in above equation we get,
$I ={I}_{0}\times {\cos}^{2}{15°}$ …(2)
We know,
$\cos { { 15 }^{ ° } } =\cfrac { \sqrt { 2+\sqrt { 3 } } }{ 2 }$
$\Rightarrow \cos ^{ 2 }{ { 15 }^{ ° }=\dfrac { 2+\sqrt { 3 } }{ 4 } }$ …(3)
Substituting equation. (3) in equation. (2) we get,
$I ={I}_{0}\times \dfrac { 2+\sqrt { 3 } }{ 4 } $
Thus, the intensity of the light transmitted by polarizer is ${ I }_{ 0 }\left( \dfrac { 2+\sqrt { 3 } }{ 4 } \right)$.

So, the correct answer is option A i.e. ${ I }_{ 0 }\left( \dfrac { 2+\sqrt { 3 } }{ 4 } \right)$.

Note:
If the $\theta$= 0 or 180°, then $I = {I}_{0}$. That is the intensity of transmitted light is maximum when transmission axes of analyzer and polarizer are parallel to each other. Initial intensity and initial after polarization remain the same.
If the $\theta$= 90°, then $I = 0$. That is the intensity of transmitted light is minimum when transmission axes of analyzer and polarizer are perpendicular to each other. The intensity of light after polarization becomes zero.