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)`)

Options

A. `I = (I_0)/8 sin^(2) 2 omega t`
B. `I = (I_0)/4 sin^(2) 2 omega t`
C. `I = (I_0)/8 cos^(2) 2 omega t`
D. `I = (I_0)/4 cos^(2) 2 omega t`

Answer

A

Explanation

Intensity after passing through `1^(st)` polaride = `(I_0)/2`
Let `2^(nd)` polaride is at angle `theta` from first polaride & `3^(rd)` polaride is at angle `90 - theta` from `2^(nd)` one then intensity after passing through `3^(rd)` polaride
` = (I_0)/2 cos^(2) theta cos^(2) (90 - theta)`
`I_("final") = (I_0)/(2) cos^(2) theta sin^(2) theta`
`(I_0)/8 sin^(2) 2 theta = (I_0)/8 sin^(2) 2 omega t`.