Malus' Law

[DeFused]

I'm a little confused on how to apply Malus' Law. Looking at the example out of TBR can someone explain how to apply it?

Vertically polarized light is sent through an empty sample cell and then into a
horizontally oriented polarizer. No light gets through the polarizer. The
experiment is repeated, this time with a dilute solution of aqueous D-Glucose in
the cell, and one-quarter of the incident light intensity gets through the polarizer.
What must be TRUE of the solution and the rotation of the polarized light?
(Assume that the solution transmits all the incident light to the polarizer.)

A. The solution is optically active and rotates the polarization by 30°.
B. The solution is optically active and rotates the polarization by 45°.
C. D-Glucose is an asymmetric molecule and the solution rotates the
polarization by 60°.
D. D-Glucose is an asymmetric molecule and the solution rotates the
polarization by 90°.

What would the initial intensity in this case?

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[sazerac]

I'm a little confused on how to apply Malus' Law. Looking at the example out of TBR can someone explain how to apply it?

Vertically polarized light is sent through an empty sample cell and then into a
horizontally oriented polarizer. No light gets through the polarizer. The
experiment is repeated, this time with a dilute solution of aqueous D-Glucose in
the cell, and one-quarter of the incident light intensity gets through the polarizer.
What must be TRUE of the solution and the rotation of the polarized light?
(Assume that the solution transmits all the incident light to the polarizer.)

A. The solution is optically active and rotates the polarization by 30°.
B. The solution is optically active and rotates the polarization by 45°.
C. D-Glucose is an asymmetric molecule and the solution rotates the
polarization by 60°.
D. D-Glucose is an asymmetric molecule and the solution rotates the
polarization by 90°.

What would the initial intensity in this case?

(1) The output of any polarized filter will always be polarized to that filter angle.
(2) The intensity of unpolarized light exiting a polarized filter will be reduced to one half the entering intensity.
(3) The intensity of polarized light exiting a polarized filter will be reduced to the cosine of (the difference between the light and the filter) times the entering intensity.

I don't have the original passage, but the answer choices only make sense if we assume that the original light is unpolarized, first passes through the vertical filter, then passes through the sugar solution which twists it some, and then passes through the horizontal filter, and they are comparing the intensity of the original unpolarized light to the light coming out of the horizontal filter. SO, assuming all that...

The light coming out of the vertical filter is oriented vertically, and reduced in intensity by one half. Then it gets rotated by some angle. Then it passes through the second horizontal filter, and gets reduced by another half. So the cosine of the angle difference at the horizontal filter must be 1/2. So the angle difference at the horizontal filter must be 60 degrees. So the sugar must rotate the light by 30 degrees.

A.

(If we are to assume the original light was always vertically polarized, then we would need to calculate the arccos(0.25) which is some number between 60 and 90 degrees, approximately 75 degrees, but 90-75=15 degrees doesn't appear to be any of the answer choices).