In Fig. 35-38, sourcesand emit long-range radio waves of wavelength400m , with the phase of the emission from ahead of that from source Bby 90° .The distance rA from Ato detector Dis greater than the corresponding distance localid="1663043743889" rBby 100m .What is the phase difference of the waves at D ?

Short Answer

Expert verified

The two waves have a 180° phase difference at the detector.

Step by step solution

01

Given data

The wavelength of the radio waves is λ=400m.

The path difference of the two waves to the detector is x=100m.

The initial phase difference of the two waves is ϕ0=90°.

02

Relation between path difference and phase difference

The phase difference of two waves of a wavelength λ having path difference x is

ϕ=2πλx ..... (1)

03

Determining the phase difference at the detector

From equation (I) the phase difference introduced in the path to the detector is,

ϕ=2π400m×100m=π2=90°

Thus, the net phase difference at the detector is

ϕD=ϕ0+ϕ=90°+90°=180°

Hence, the required phase difference is 180°.

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Most popular questions from this chapter

In the double-slit experiment of Fig. 35-10, the electric fields of the waves arriving at point P are given by

E1=(2.00μV/m)sin[1.26×1015t]E2=(2.00μV/m)sin[1.26×1015t+39.6rad]

Where, timetis in seconds. (a) What is the amplitude of the resultant electric field at point P ? (b) What is the ratio of the intensity IPat point P to the intensity Icenat the center of the interference pattern? (c) Describe where point P is in the interference pattern by giving the maximum or minimum on which it lies, or the maximum and minimum between which it lies. In a phasor diagram of the electric fields, (d) at what rate would the phasors rotate around the origin and (e) what is the angle between the phasors?

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Reflection by thin layers. In Fig. 35-42, light is incident perpendicularly on a thin layer of material 2 that lies between (thicker) materials 1 and 3. (The rays are tilted only for clarity.) The waves of rays r1and r2interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35- 2 refers to the indexes of refraction n1,n2and n3, the type of interference, the thin-layer thickness in nanometres, and the wavelength λ in nanometres of the light as measured in air. Where is missing, give the wavelength that is in the visible range. Where is missing, give the second least thickness or the third least thickness as indicated.

Figure 35-56a show two light rays that are initially in phase as they travel upward through a block of plastic, with wavelength 400 nm as measured in air. Light ray r1exits directly into air. However, before light ray r2exits into air, it travels through a liquid in a hollow cylinder within the plastic. Initially the height Lliqof the liquid is 40.0 μm, but then the liquid begins to evaporate. Let θbe the phase difference between raysr1and r2once they both exit into the air. Figure 35-56b, showsθ versus the liquid’s heightLliq until the liquid disappears, with θgiven in terms of wavelength and the horizontal scale set byLs=40.00μm.What are (a) the index of refraction of the plastic and (b) the index of refraction of the liquid?

Figure 35-28 shows four situations in which light reflects perpendicularly from a thin film of thickness L sandwiched between much thicker materials. The indexes of refraction are given. In which situations does Eq. 35-36 correspond to the reflections yielding maxima (that is, a bright film).

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