Question

Unpolarized light of intensity $32{\text{Wm}}^{-2}$ passes through three polarizers such that transmission axes of the first and second polarizer makes an angle ${30}^{\circ }$ with each other and the transmission axis of the last polarizer makes an angle $\theta$ with the axis of second polarizer as shown in figure. The intensity of final emerging light will be-

• A. $32{\text{Wm}}^{-2}$
• B. $16{\text{Wm}}^{-2}$
• C. $12{\text{Wm}}^{-2}$
• D. $3{\text{Wm}}^{-2}$

Answer 1

The correct option is D $3{\text{Wm}}^{-2}$



Let, $P_1,P_2$ and $P_3$ be the three polarisers, and angle between $P_1$ and $P_2$=${30}^{\circ }$

And angle between $P_2$ and $P_3$ is,

$\theta ={90}^{\circ }-{30}^{\circ }={60}^{\circ }$

$I_0=32{\text{Wm}}^{-2}$

The intensity of light transmitted by $P_1$ is,

$I_1=\frac{I_0}{2}=\frac{32}{2}=16{\text{Wm}}^{-2}$

According to Malu's law, the intensity of light transmitted is given by,

$I=I_0{\cos }^2\theta$

Hence, intensity of light emerging out from the $P_2$ is,

$I_2=I_1{\cos }^2{30}^{\circ }=16{\left(\sqrt{\frac{\sqrt{3}}{2}}\right)}^2=12{\text{Wm}}^{-2}$

Similarly, intensity of light emerges out from the $P_2$ is,

$I_3=I_2{\cos }^2{60}^{\circ }=12{\left(\frac{1}{2}\right)}^2=3{\text{Wm}}^{-2}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.