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 \(\frac{1}{4}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Calculate the frequency of an EM wave with a wavelength the size of (a) the thickness of a piece of paper \((60 \mu \mathrm{m}),(\mathrm{b})\) a \(91-\mathrm{m}\) -long soccer field, (c) the diameter of Earth, (d) the distance from Earth to the Sun.
The index of refraction of water is \(1.33 .\) (a) What is the speed of light in water? (b) What is the wavelength in water of a light wave with a vacuum wavelength of \(515 \mathrm{nm} ?\)
An electric dipole antenna used to transmit radio waves is oriented vertically.At a point due north of the transmitter, how should a magnetic dipole antenna be oriented to serve as a receiver?
A polarized beam of light has intensity \(I_{0} .\) We want to rotate the direction of polarization by \(90.0^{\circ}\) using polarizing sheets. (a) Explain why we must use at least two sheets. (b) What is the transmitted intensity if we use two sheets, each of which rotates the direction of polarization by \(45.0^{\circ} ?\) (c) What is the transmitted intensity if we use four sheets, each of which rotates the direction of polarization by \(22.5^{\circ} ?\)
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.
See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free