Two polarizers A and B are aligned so that their transmission axes are vertical and horizontal, respectively. A third polarizer is placed between these two with its axis aligned at angle theta with respect to the vertical.
Assuming vertically polarized light of intensity Io is incident upon polarizer A, find an expression for the light intensity I transmitted through this three-polarizer sequence.
The answer is: I=1/4*Io*sin^2(2*theta)
Please let me know how to get the answer. Thanks.
Assuming vertically polarized light of intensity Io is incident upon polarizer A, find an expression for the light intensity I transmitted through this three-polarizer sequence.
The answer is: I=1/4*Io*sin^2(2*theta)
Please let me know how to get the answer. Thanks.
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Hi Wicky
The transmission through the second filter is given by Malus' Law
I1= Io cos^2 (theta)
The angle between the intermediate filter and B is (90 - theta) and the transmission through filter B will
be !2 = !1 cos^2 (90 - theta)
= Io cos^2 (theta) cos^2 (90 - theta)
now cos (90- theta) = cos 90 cos theta - sin 90 sin theta = sin theta (compound angles)
so cos^2(90-thate) = sin^2 (theta)
so I2 = Io cos^2 (theta) sin^2 (theta)
now sin ( 2 theta) = 2 sin (theta) cos (theta) (double angles)
so sin^2( 2 theta) = 4 sin^2(theta) cos^2(theta)
so 1/4 * sin^2 (2 theta) = sin^2(theta)cos^2(theta)
hence I2 = 1/4* sin^2 (2 theta)
(The references in brackets are to trig identities which you learn in A2 Maths given that name ie 'double angle identities' etc - you don't do these in Physics!!)
The transmission through the second filter is given by Malus' Law
I1= Io cos^2 (theta)
The angle between the intermediate filter and B is (90 - theta) and the transmission through filter B will
be !2 = !1 cos^2 (90 - theta)
= Io cos^2 (theta) cos^2 (90 - theta)
now cos (90- theta) = cos 90 cos theta - sin 90 sin theta = sin theta (compound angles)
so cos^2(90-thate) = sin^2 (theta)
so I2 = Io cos^2 (theta) sin^2 (theta)
now sin ( 2 theta) = 2 sin (theta) cos (theta) (double angles)
so sin^2( 2 theta) = 4 sin^2(theta) cos^2(theta)
so 1/4 * sin^2 (2 theta) = sin^2(theta)cos^2(theta)
hence I2 = 1/4* sin^2 (2 theta)
(The references in brackets are to trig identities which you learn in A2 Maths given that name ie 'double angle identities' etc - you don't do these in Physics!!)