Question

Two polarizing sheets have their transmission axes crossed so that no light is transmitted. A third sheet is inserted so that its transmission axis makes an angle $\theta$ with the transmission axis of the first sheet. If the middle polarizing sheet is rotating at an angular speed $\omega$ about an axis parallel with the light beam, find an expression for the final intensity transmitted through after all three sheets as a function of time. (Original intensity was $I_0$ )

• A. $I=\frac{I_0}{8}si{n}^{2}2\omega t$
• B. $I=\frac{I_0}{4}si{n}^{2}2\omega t$
• C. $I=\frac{I_0}{8}co{s}^{2}2\omega t$
• D. $I=\frac{I_0}{4}co{s}^{2}2\omega t$

Answer 1

Answer: (A)

$=\frac{I_0}{2}{\cos }^2\theta {\cos }^2(90-\theta )$ Intensity after passing through $I^{\text{st }}$ polaride $=\frac{I_0}{2}$ Let $2^{\text{nd }}$ polaride is at angle $\theta$ from first polaride and $3^{\text{rd }}$ polaride is at angle $90-\theta$ from $2^{\text{nd }}$ one then intensity after passing through
$2^{\text{nd }}$ polaride $=\frac{I_0}{2}{\cos }^2\theta$
Intensity after passing through $3^{\text{rd }}$ polaride
$\begin{array}{l}{I}_{\text{final }}=\frac{I_0}{2}{\cos }^2\theta {\sin }^2\theta \\ =\frac{I_0}{8}{\sin }^{2}2\theta =\frac{I_0}{8}{\sin }^{2}2\omega t\end{array}$