Light is entering a polarizer or set of polarizers from the left. Determine the

Question

Light is entering a polarizer or set of polarizers from the left. Determine the relative intensity of the light exiting on the right after passing through all of the polarizers. The dashed vertical lines represent the vertical reference. The solid lines represent either the polarization angle of the light or the transmission angle of the polarizer. CW represents clockwise and CCW represents counterclockwise as viewed from the right looking back to the left along the path of the light. Unpolarized light enters from the left and passes first through a polarizer oriented 45°CW from the vertical and the through a polarizer oriented 45° CCW. Vertically polarized light passes through a polarizer with the transmission axis oriented 90° relative to the vertical and then through a second polarizer oriented vertically. Vertically polarized light enters from the left and passes first through a polarizer oriented 45°CW from the vertical and the through a polarizer oriented 45° CW. Horizontally polarized light passes through a polarizer with the transmission axis oriented 90° relative to the vertical and then through a second polarizer oriented 90° relative to the vertical. Submit Answer Incorrect. Tries 1/10 Previous Tries This discussion is closed.

Explanation / Answer

key concept

1 ) when unpolarized passes through first polarizer then intensity became 1/2 and then apply malus law.

2) if the light is polarized then directly apply malus law.

so )

part) a

unpolarized light

suppose light intensity was Io initially in all parts

after first polarizer

i = Io /2

after second polarizer I = i *[cos( theta) ]^2 = Io/2* cos( 45 --45) = 0 answer

Part b)

Vertically polarized light

I = i *[cos( theta) ]^2

= i *cos( 90) ^2 = 0

so from also second polarizer no intesity will pass so fianl intensity = 0 answer

part c)

I = Io *[cos( theta) ]^2

= Io *cos( 45) ^2 = 0.5*Io   CW

from second plarizer  45° CW

I = 0.5 *Io*[cos( theta) ]^2 = 0.5 *0.5* Io = 0.25Io ..........answer

part d

I = Io *[cos( theta) ]^2

= Io *cos( 0) ^2 = Io   90° relative to vertical

from 2 polarizer

I = Io *[cos( theta) ]^2

= Io *cos( 0 ) ^2 = Io .................answer

Related Questions

Navigate

• Browse (All)
• Browse L
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.