## Mastering Physics Solutions: Polarization of Light and Malus’s Law

Polarization of Light and Malus’s Law

Part A = θTA – θ0
Part B = θTA
Part C = 18.7 W/m2
Part D = I0 = 2I = 5.81760
Part E = I2 = 0I0 = 0
Part F = See below

Solution Below:

Part A

A beam of polarized light with intensity I0 and polarization angle θ0 strikes a polarizer with transmission axis θTA. What angle θ should be used in Malus’s law to calculate the transmitted intensity I1?

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.

θTA – θ0

θTA – θ0

Part B

What is the polarization angle θ1 of the light emerging from the polarizer?

θTA

θTA

Part C

If I0 = 20.0 W/m2, θ0 = 25.0 degrees, and θTA = 40.0 degrees, what is the transmitted intensity I1?
Express your answer numerically in watts per square meter.

Just use the formula:

I1 = I0 * cos(π * (θTA – θ0) / 180)2

The answer is 18.7 W/m2

18.7 W/m2

Part D

One way to produce a beam of polarized light with intensity I and polarization angle θ would be to pass unpolarized light with intensity I0 through a polarizer whose transmission axis is oriented such that θTA = θ. How large must I0 be if the transmitted light is to have intensity I?
Express your answer as a decimal number times the symbol I. For example, if I0 = (1/4)I, enter 0.25 * I.

The answer is I0 = 2I = 5.81760

I0 = 2I = 5.81760

Part E

A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polarizer with transmission axis θTA,2, where θTA,2 – θTA,1 = 90 degrees (in other words the two axes are perpendicular to one another). What is the intensity I2 of the light beam emerging from the second polarizer?
Express your answer as a decimal number times the symbol I0. For example, if I2 = (1/4)I0, enter 0.25 * I_0.

The answer is I2 = 0I0 = 0

I2 = 0I0 = 0

Part F

Notice that a polarizer modifies the light intensity according to Malus’s law and also reorients the polarization angle of the beam to match its own transmission axis. Hence it is possible for light to pass through a pair of crossed polarizers if a third polarizer is inserted between them with an intermediate transmission axis direction. What is the new intensity I2 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 I0. For example, if I2 = (1/4)I0, enter 0.25 * I_0.

The following formula gives the answer:

I2 = 0.5(cos(45π / 180))2 (cos(45π / 180))2 I0 = 3.44*10-2

Or

0.125I0 = 9.50*10-2

See above

Tagged with:

### 2 Responses to Mastering Physics Solutions: Polarization of Light and Malus’s Law

1. seri says:

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?

• Mastering Physics Solutions says:

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