/*! 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} Q21E Consider two antennas separated ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Consider two antennas separated by 9.00m that radiate in phase at 120MHz, as described in Exercise . A receiver placed 150m from both antennas measures an intensity I0.The receiver is moved so that it is closer to one antenna than to the other. (a) What is the phase differenceϕbetween the two radio waves produced by this path difference? (b) In terms of I0, what is the intensity measured by the receiver at its new position?

Short Answer

Expert verified
  1. The phase difference ϕbetween the two radio waves produced by this path difference is 4.52rad .
  2. The intensity measured by the receiver at its new position is 0.404I0.

Step by step solution

01

formulas used to solve the question

Intensity is given by

I=I0cos2(ϕ2) (1)

Phase difference is given by

ϕ=∆r*2πλ (2)

02

Calculate the phase difference

Given:

d = 9.0m

f=120MHz=120*106Hz

r1=r2=150m∆r2=1.8m

The speed of electromagnetic waves is as same as the speed of light. So, the speed of wave emitted from both stations is given by

c=λf⇒λ=cf (3)

In equation (2), plug equation (3),

ϕ=2πf∆rc (4)

When the receiver is moved closer to one antenna than to the other,

Plug the given in equation (4),

Ï•=2Ï€*120*106*1.803.0*108=4.52rad

03

Calculate the intensity

From equation (1), plug the given

I=I0cos2(4.522)=0.404I0

Thus, the phase differenceϕ between the two radio waves produced by this path difference is 4.52rad. The intensity measured by the receiver at its new position is 0.404I0.

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!

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

.. DATA While researching the use of laser pointers, you conduct a diffraction experiment with two thin parallel slits.

Your result is the pattern of closely spaced bright and dark fringes shown in Fig. P36.63. (Only the central portion of the pattern is shown.) You measure that the bright spots are equally spaced at centre to centre (except for the missing spots) on a screen that is from the slits. The light source was a helium–neon laser producing a wavelength of . (a) How far apart are the two slits? (b) How wide is each one?

34.15 The thin glass shell shown in Fig. E34.15 has a spherical shape with a radius of curvature of 12cm, and both of its surfaces can act as mirrors. A seed high is placed 15.0cmfrom the center of the mirror along the optic axis, as shown in the figure. (a) Calculate the location and height of the image of this seed. (b) Suppose now that the shell is reversed. Find the location and height of the seed’s image.

(a) Prove that when two thin lenses with focal lengths f1and f2are placed in contact, the focal length Æ’ of the combination is given by the relationship 1f=1f1+1f2 (b) A converging meniscus lens (see Fig. 34.32a) has an index of refraction of 1.55 and radii of curvature for its surfaces of magnitudes 4.50 cm and 9.00 cm. The concave surface is placed upward and filled with carbon tetrachloride (CCI4), which has n = 1.46. What is the focal length of the CCI4-glass combination?

The left end of a long glass rod 8.00 cm in diameter, with an index of refraction of 1.60, is ground and polished to a convex hemispherical surface with a radius of 4.00 cm. An object in the form of an arrow 1.50 mm tall, at right angles to the axis of the rod, is located on the axis 24.0 cm to the left of the vertex of the convex surface. Find the position and height of the image of the
arrow formed by paraxial rays incident on the convex surface. Is the image erect or inverted?

The eyes of amphibians such as frogs have a much flatter cornea but a more strongly curved (almost spherical) lens than do the eyes of air-dwelling mammals. In mammalian eyes, the shape (and therefore the focal length) of the lens changes to enable the eye to focus at different distances. In amphibian eyes, the shape of the lens doesn’t change. Amphibians focus on objects at different distances by using specialized muscles to move the lens closer to or farther from the retina, like the focusing mechanism of a camera. In air, most frogs are near-sighted; correcting the distance vision of a typical frog in air would require contact lenses with a power of about -6.0 D .Given that frogs are nearsighted in air, which statement is most likely to be true about their vision in water? (a) They are even more nearsighted; because water has a higher index of refraction than air, a frog’s ability to focus light increases in water. (b) They are less nearsighted, because the cornea is less effective at refracting light in water than in air. (c) Their vision is no different, because only structures that are internal to the eye can affect the eye’s ability to focus. (d) The images projected on the retina are no longer inverted, because the eye in water functions as a diverging lens rather than a converging lens.

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.