Light of intensity I0 is polarized vertically and is incident on an analyzer rotated at an angle ? from the vertical. Find the angle ? if the transmitted light has intensity I = (0.750)I0, I = (0.500)I0, I = (0.250)I0, and I = 0. (Enter your answers in degrees.)
(a) I = (0.750)I0 °
(b) I = (0.500)I0 °
(c) I = (0.250)I0 °
(d) I = 0 °
theta = A
when polarized light passes thorugh polarizer, Intensity will be according to Malus's law:
I = I0*(cos A)^2
Cos A = sqrt (I/I0)
A = arccos sqrt (I/I0)
Part (a),
when given that, I = 0.750*I0
I/I0 = 0.750
So, Angle will be
A = arccos (sqrt 0.750) = 30 deg
Part (b),
when given that, I = 0.500*I0
I/I0 = 0.500
So, Angle will be
A = arccos (sqrt 0.500) = 45 deg
Part (c)
when given that, I = 0.250*I0
I/I0 = 0.250
So, Angle will be
A = arccos (sqrt 0.250) = 60 deg
Part (d),
when given that, I = 0*I0
I/I0 = 0
So, Angle will be
A = arccos (sqrt 0) = 90 deg
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