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Chapter 22: Problem 45

Unpolarized light passes through two polarizers in turn with polarization axes at \(45^{\circ}\) to one another.What is the fraction of the incident light intensity that is transmitted?

Short Answer

Expert verified
Answer: \(\frac{1}{4}\)

Step by step solution

01

Recall Malus' law

Malus' law states that the intensity of the transmitted light through a polarizer can be calculated by the formula: \(I_t = I_i \cdot \cos^2{\theta}\), where \(I_t\) is the transmitted intensity, \(I_i\) is the initial intensity, and \(\theta\) is the angle between the polarization axis of the polarizer and the initial polarization direction.
02

Unpolarized light through the first polarizer

When unpolarized light passes through a polarizer, half the unpolarized light gets absorbed, and the light gets polarized with the remaining half intensity. So, after passing through the first polarizer, the intensity becomes \(I' = \frac{1}{2}I\), where \(I\) is the initial incident intensity and \(I'\) is the intensity after the first polarizer.
03

Polarized light through the second polarizer

Now the light is polarized and passes through the second polarizer with the axis at \(45^{\circ}\) to the first polarizer's axis. To find the transmitted intensity after the second polarizer, we use Malus' law: \(I_t = I' \cdot \cos^2{45^{\circ}}\) Recalling that \(\cos{45^{\circ}} = \frac{1}{\sqrt{2}}\), we substitute this into the equation: \(I_t = I' \cdot \left(\frac{1}{\sqrt{2}}\right)^2\)
04

Finding the fraction of transmitted intensity relative to initial intensity

Replace \(I'\) in the above equation with \(\frac{1}{2}I\): \(I_t = \left(\frac{1}{2}I\right) \cdot \left(\frac{1}{2}\right)\) \(I_t = \frac{1}{4}I\) To find the fraction of the transmitted intensity with respect to the initial intensity, divide \(I_t\) by \(I\): Fraction \(= \frac{I_t}{I} = \frac{\frac{1}{4}I}{I} = \frac{1}{4}\) So, the fraction of the incident light intensity that is transmitted through the two polarizers is \(\frac{1}{4}\).

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Most popular questions from this chapter

Light polarized in the \(x\) -direction shines through two polarizing sheets. The first sheet's transmission axis makes an angle \(\theta\) with the \(x\) -axis and the transmission axis of the second is parallel to the \(y\) -axis. (a) If the incident light has intensity \(I_{0},\) what is the intensity of the light transmitted through the second sheet? (b) At what angle \(\theta\) is the transmitted intensity maximum?
An AM radio station broadcasts at \(570 \mathrm{kHz}\). (a) What is the wavelength of the radio wave in air? (b) If a radio is tuned to this station and the inductance in the tuning circuit is \(0.20 \mathrm{mH},\) what is the capacitance in the tuning circuit? (c) In the vicinity of the radio, the amplitude of the electric field is \(0.80 \mathrm{V} / \mathrm{m} .\) The radio uses a coil antenna of radius \(1.6 \mathrm{cm}\) with 50 turns. What is the maximum emf induced in the antenna, assuming it is oriented for best reception? Assume that the fields are sinusoidal functions of time.
What is the wavelength of the radio waves broadcast by an FM radio station with a frequency of 90.9 MHz?
In musical acoustics, a frequency ratio of 2: 1 is called an octave. Humans with extremely good hearing can hear sounds ranging from \(20 \mathrm{Hz}\) to \(20 \mathrm{kHz}\), which is approximately 10 octaves (since $2^{10}=1024 \approx 1000$ ). (a) Approximately how many octaves of visible light are humans able to perceive? (b) Approximately how many octaves wide is the microwave region?
How far does a beam of light travel in 1 ns?
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