A beam of polarized light is sent into a system of two polarizing sheets. Relative to the polarization direction of that incident light, the polarizing directions of the sheets are at angles θfor the first sheet and 90°for the second sheet. If 0.10of the incident intensity is transmitted by the two sheets, what is θ?

Short Answer

Expert verified

Polarizing direction of the first sheet θ=20°.

Step by step solution

01

Step 1: Identification of the given data

  1. Polarizing direction of the second sheet 90°.
  2. The fraction of intensity transmitted by the two sheets is 0.10.
02

Determining the concept

Cosine squared law states that when a plane of polarized light is passed through an analyzer, the intensity of the transmitted beam from the analyzer is directly propositional to the cosine of the angle between the transmission axes of the polarizer and the analyzer.

The formula is as follows:

I=I0cos2θ

where,

I is the radiant intensity,

cosθis the angle between the direction of the incident light and the surface normal,

03

Determining the polarizing direction of the first sheet  θ

The angle of incidence of the beam with the second sheet is 90°.

The second angle is as follows:

θ2=90θ

The intensity of light fromthefirst polarizing sheet is as follows:

I1=I0cos2θ

The intensity of light fromthesecond polarizing sheet is as follows:

I2=I1cos2θ2

Now plug the value I1 and θ2 in the above equation.

I2=I0cos2θcos2(90θ)I2=I0cos2θsin2θI2=I014sin2(2θ)

Since the intensity of the transmitted beam is 0.1of the incident beam,

I2=0.1I0

Substitute all the value in the above equation.

I014sin2(2θ)=0.10I014sin2(2θ)=0.10sin2(2θ)=0.10×4sin2(2θ)=0.4sin(2θ)=0.63245(2θ)=sin-1(0.63245)2θ=40°θ=20°

Hence, polarizing direction of the first sheet θ=20° .

The value of the required angle can be calculated using the cosine squared law.

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

An unpolarized beam of light is sent into a stack of four polarizing sheets, oriented so that the angle between the polarizing directions of adjacent sheets is30°. What fraction of the incident intensity is transmitted by the system?

Question: What is the radiation pressure 1.5maway from a 500wlight bulb? Assume that the surface on which the pressure is exerted faces the bulb and is perfectly absorbing and that the bulb radiates uniformly in all directions.

Figure 33-28 shows the electric and magnetic fields of an electromagnetic wave at a certain instant. Is the wave traveling into the page or out of the page?

In Fig. 33-65, a light ray enters a glass slab at point A at an incident angle θ1=45.0°and then undergoes total internal reflection at point B. (The reflection at A is not shown.) What minimum value for the index of refraction of the glass can be inferred from this information?

Question: Frank D. Drake, an investigator in the SETI (Search for Extra-Terrestrial Intelligence) program, once said that the large radio telescope in Arecibo, Puerto Rico (Fig.33-36), “can detect a signal which lays down on the entire surface of the earth a power of only one picowatt.” (a) What is the power that would be received by the Arecibo antenna for such a signal? The antenna diameter is 300m.(b) What would be the power of an isotropic source at the center of our galaxy that could provide such a signal? The galactic center is 2.2×104lyaway. A light-year is the distance light travels in one year.

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