/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Unpolarized light is incident on... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Unpolarized light is incident on two ideal polarizing filters. The second filter's axis is rotated through an angle \(\theta\) relative to that of the first filter. If the intensity of light emerging from the second filter is \(1 / 10\) the intensity of the incident light, what is \(\theta ?\)

Short Answer

Expert verified
\( \theta \approx 63.43^\circ \).

Step by step solution

01

Understand the Situation

We have unpolarized light passing through two polarizing filters. The axis of the second filter is rotated by an angle \( \theta \) with respect to the axis of the first filter. We need to find this angle \( \theta \).
02

Intensity of Light After First Filter

When unpolarized light passes through the first polarizer, its intensity is reduced to half. If the initial intensity is \( I_0 \), the intensity after the first filter is \( I_1 = \frac{1}{2}I_0 \).
03

Apply Malus's Law for Second Filter

Malus's Law states that when polarized light passes through a second polarizer, the transmitted intensity \( I \) is related to the initial intensity \( I_1 \) and the angle \( \theta \) between the polarizer's axes: \[ I = I_1 \cos^2\theta \] Where \( I_1 \) is the intensity after the first filter.
04

Set Up the Equation for Final Intensity

We know that the final intensity leaving the second filter is \( \frac{1}{10} I_0 \). Set up the equation: \[ \frac{1}{10} I_0 = \left( \frac{1}{2} I_0 \right) \cos^2\theta \] This equation relates initial and final intensities with the angle \( \theta \).
05

Solve for \(\cos^2\theta\)

Divide both sides of the equation \[ \frac{1}{10} I_0 = \frac{1}{2} I_0 \cos^2\theta \]by \( \frac{1}{2} I_0 \):\[ \cos^2\theta = \frac{1}{10} \times \frac{2}{1} = \frac{1}{5} \].
06

Calculate \(\theta\)

To find \( \theta \), take the square root of both sides of \( \cos^2\theta = \frac{1}{5} \):\[ \cos\theta = \sqrt{\frac{1}{5}} \].Now, use the inverse cosine function to find \( \theta \):\[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{5}} \right) \].
07

Final Answer

Calculate \( \theta \) using a calculator to find the angle. Approximate \( \theta \approx 63.43^\circ \).

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 91Ó°ÊÓ!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Malus's Law
Malus's Law is a fundamental principle in optics that describes how light intensity changes as it passes through polarizing filters. When polarized light encounters a second polarizer, the transmitted intensity is determined by the angle between the incoming light's polarization direction and the polarizer's axis. According to Malus's Law, the transmitted intensity \( I \) is given by the equation:\[ I = I_0 \cos^2 \theta \]where \( I_0 \) is the initial intensity of the polarized light and \( \theta \) is the angle between the light's polarization direction and the polarizer's axis.
This law is crucial for understanding how varying the angle \( \theta \) affects the amount of light passing through multiple polarizers. By applying Malus's Law, one can predict and calculate the light intensity at any given angle. It's particularly useful in scenarios where light manipulation is required, such as in photography and LCD technology.
Intensity of Light
Intensity of light refers to the amount of energy that passes through a given area in a given time. In the context of polarizing filters, intensity significantly changes based on the orientation and number of filters used.
When unpolarized light is incident on a polarizer, its intensity is reduced. Initially, the intensity of unpolarized light \( I_0 \) gets halved as it passes through the first polarizer since only the component aligned with the filter’s axis is transmitted. Hence, the intensity after the first polarizer becomes \( \frac{1}{2} I_0 \).
As the light then passes through a second polarizer, the intensity is further modified, which is dictated by Malus's Law. In situations where multiple polarizers are used, understanding how the intensity changes is essential for calculating the light emerging from the system. This helps engineers and scientists in designing optical systems for various applications.
Polarizing Filters
Polarizing filters are optical devices that allow light waves of a specific polarization to pass through while absorbing or reflecting others. They are essential tools in controlling and analyzing light properties in numerous technical fields.
The first polarizing filter affects unpolarized light by allowing only the component of light waves aligned with its axis to pass. This process results in polarized light, which has reduced intensity, specifically halved compared to the initial intensity of unpolarized light.
The second polarizer, when placed at an angle relative to the first, further modifies the intensity based on Malus's Law. By strategically rotating the second filter’s axis, one can control the intensity of light. This capability is leveraged in photography to reduce glare and enhance image quality, and in scientific applications where precise control of light intensity is needed.
Unpolarized Light
Unpolarized light is light in which the electric field vibrates in all possible directions perpendicular to the direction of propagation. Most common light sources, such as the sun or incandescent bulbs, emit unpolarized light.
When unpolarized light passes through a polarizing filter, it becomes polarized, meaning its electric field vibrates in only one direction.
This transformation is crucial for various optical instruments and technologies, as polarizing filters can control the intensity and direction of light and thereby improve visual clarity. Understanding the nature of unpolarized light and its behavior when interacting with polarizing materials is fundamental for both basic physics studies and sophisticated technological applications.

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

A beam of light in air makes an angle of \(47.5^{\circ}\) with the surface (not the normal) of a glass plate having a refractive index of 1.66 . (a) What is the angle between the reflected part of the beam and the surface of the glass? (b) What is the angle between the refracted beam and the surface (not the normal) of the glass?

A small helium-neon laser emits red visible light with a power of \(3.20 \mathrm{~mW}\) in a beam that has a diameter of \(2.50 \mathrm{~mm}\). (a) What are the amplitudes of the electric and magnetic fields of the light? (b) What are the average energy densities associated with the electric field and with the magnetic field? (c) What is the total energy contained in a \(1.00 \mathrm{~m}\) length of the beam?

A sinusoidal electromagnetic wave has an average intensity of \(100 \mathrm{~W} / \mathrm{m}^{2} .\) By what factor would the electric-field amplitude of the wave have to be increased in order for the wave to have an average intensity of \(10,000 \mathrm{~W} / \mathrm{m}^{2} ?\)

A geostationary communications satellite orbits the earth directly above the equator at an altitude of \(35,800 \mathrm{~km}\). Calculate the time it would take a cell phone signal to travel from a point on the equator to the satellite and back. Would this delay be noticeable in a conversation?

The indices of refraction for violet light \((\lambda=400 \mathrm{nm})\) and red light \((\lambda=700 \mathrm{nm})\) in diamond are 2.46 and \(2.41,\) respectively. A ray of light traveling through air strikes the diamond surface at an angle of \(53.5^{\circ}\) to the normal. Calculate the angular separation between these two colors of light in the refracted ray.

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.