After passing through a polariser a linearly polarised light of intensity I is incident on an analyser making an angle of 30° with that of the polariser. The intensity of light emitted from the analyser will be :

Question

After passing through a polariser a linearly polarised light of intensity I is incident on an analyser making an angle of 30° with that of the polariser. The intensity of light emitted from the analyser will be : (A) ( I /2) (B) (I/3) (C) (3 I/4) (D) (2 I/3)

Tardigrade Admin · Accepted Answer

According to Malus law I 2= I 1 cos 2 θ Where I 1 is incident Polarised light I 2= I cos 2 30= I ((√3/2))2 I 2=(3 I /4)

Q. After passing through a polariser a linearly polarised light of intensity $I$ is incident on an analyser making an angle of $3 0^{\circ}$ with that of the polariser. The intensity of light emitted from the analyser will be :

$\frac{I}{2}$

$\frac{I}{3}$

$\frac{3 I}{4}$

$\frac{2 I}{3}$

According to Malus law
$I_{2} = I_{1} cos^{2} θ$
Where $I_{1}$ is incident Polarised light
$I_{2} = I cos^{2} 30 = I (\frac{3}{2})^{2}$
$I_{2} = \frac{3 I}{4}$

Q. After passing through a polariser a linearly polarised light of intensity I is incident on an analyser making an angle of 30∘ with that of the polariser. The intensity of light emitted from the analyser will be :

2I​

3I​

43I​

32I​