**Polarization of Light and Malus’s Law**

Part A = θ_{TA} – θ_{0}

Part B = θ_{TA}

Part C = 18.7 W/m^{2}

Part D = I_{0} = 2I = 5.81760

Part E = I_{2} = 0I_{0} = 0

Part F = See below

**Solution Below:**

Part A

_{0}and polarization angle θ

_{0}strikes a polarizer with transmission axis θ

_{TA}. What angle θ should be used in Malus’s law to calculate the transmitted intensity I

_{1}?

This process is illustrated in the figure, where the polarization of the light wave is visually illustrated by a magenta double arrow oriented in the direction of polarization, the transmission axis of the polarizer is represented by a blue double arrow, and the direction of motion of the wave is illustrated by a purple arrow.

The answer is:

θ_{TA} – θ_{0}

θ_{TA} – θ_{0}

Part B

_{1}of the light emerging from the polarizer?

The answer is:

θ_{TA}

θ_{TA}

Part C

_{0}= 20.0 W/m

^{2}, θ

_{0}= 25.0 degrees, and θ

_{TA}= 40.0 degrees, what is the transmitted intensity I

_{1}?

Express your answer numerically in watts per square meter.

Just use the formula:

I_{1} = I_{0} * cos(π * (θ_{TA} – θ_{0}) / 180)^{2}

The answer is 18.7 W/m^{2}

18.7 W/m^{2}

Part D

_{0}through a polarizer whose transmission axis is oriented such that θ

_{TA}= θ. How large must I

_{0}be if the transmitted light is to have intensity I?

Express your answer as a decimal number times the symbol I. For example, if I

_{0}= (1/4)I, enter 0.25 * I.

The answer is I_{0} = 2I = 5.81760

I_{0} = 2I = 5.81760

Part E

_{0}falls first upon a polarizer with transmission axis θ

_{TA,1}then upon a second polarizer with transmission axis θ

_{TA,2}, where θ

_{TA,2}– θ

_{TA,1}= 90 degrees (in other words the two axes are perpendicular to one another). What is the intensity I

_{2}of the light beam emerging from the second polarizer?

Express your answer as a decimal number times the symbol I

_{0}. For example, if I

_{2}= (1/4)I

_{0}, enter 0.25 * I_0.

The answer is I_{2} = 0I_{0} = 0

I_{2} = 0I_{0} = 0

Part F

_{2}of the light emerging from the final polarizer in Part E if a third polarizer (Polarizer A in the figure), whose transmission axis is offset 45 degrees from each of the others, is inserted between the original two polarizers?

Express your answer as a decimal number times the symbol I

_{0}. For example, if I

_{2}= (1/4)I

_{0}, enter 0.25 * I_0.

The following formula gives the answer:

I_{2} = 0.5(cos(45π / 180))^{2} (cos(45π / 180))^{2} I_{0} = 3.44*10^{-2}

Or

0.125I_{0} = 9.50*10^{-2}

See above

A polaroid sheet and an analyzer are placed such that their transmission axes are

parallel. The analyzer is then rotated by 22.5 o . What is the intensity of the

transmitted light as a fraction of its original value?

The intensity of the of the transmitted light can be calculated using the formula I(t) = I(0)*cos^2(θ). For your example, the transmitted light should have an intensity of about 0.8536 : 1, or 85.36% of the original