In Fig. 35-37, two radio frequency point sources S1and S2, separated by distance d=2.0m, are radiating in phase with λ=0.50m. A detector moves in a large circular path around the two sources in a plane containing them. How many maxima does it detect?

Short Answer

Expert verified

The total number of the maxima is 16.

Step by step solution

01

Write the given data from the question

The distance between the slits, d=2m.

The wavelength, λ=0.50m.

02

Determine the formulas to calculate the number of the maxima detect by the detector

The condition for the maxima in Young’s experiment is given as follows.

dsinθ=mλ …… (1)

Here,d is the distance between the slits, λis the wavelength, mis the order, andθis the angular separation.

03

Calculate the number of the maxima detect by the detector

Calculate the number of the maxima.

Substitute 0.50mfor λand 2m for d into equation (1).

2sinθ=m×0.50sinθ=12×m×0.50sinθ=14×m

Now,

sinθ⩽1m4⩽1.

The value of m can be negative as well as positive. Therefore, m will have -4,-3,-2,-1,0,+1,+2,+3,+4.

For each of the value there is two values of the θ.

Therefore, one value for -900Cand other for +900C.

Hence, the total number of the maxima is 16.

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Most popular questions from this chapter

A thin film with index of refraction n=1.40 is placed in one arm of a Michelson interferometer, perpendicular to the optical path. If this causes a shift of 7.0 bright fringes of the pattern produced by light of wavelength 589nm, what is the film thickness?

Transmission through thin layers. In Fig. 35-43, 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.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3andr4interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refractionn1,n2andn3the type of interference, the thin-layer thicknessLin nanometers, and the wavelengthλin nanometers of the light as measured in air. Whereλis missing, give the wavelength that is in the visible range. WhereLis missing, give the second least thickness or the third least thickness as indicated.

Transmission through thin layers. In Fig. 35-43, 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.) Part of the light ends up in material 3 as ray r3(the light does not reflect inside material 2) and r4(the light reflects twice inside material 2). The waves of r3and r4interfere, and here we consider the type of interference to be either maximum (max) or minimum (min). For this situation, each problem in Table 35-3 refers to the indexes of refraction n1,n2and n3, the type of interference, the thin-layer thickness L in nanometers, and the wavelength in nanometers of the light as measured in air. Where λis missing, give the wavelength that is in the visible range. Where Lis missing, give the second least thickness or the third least thickness as indicated.

Figure 35-24a gives intensity lversus position x on the viewing screen for the central portion of a two-slit interference pattern. The other parts of the figure give phasor diagrams for the electric field components of the waves arriving at the screen from the two slits (as in Fig. 35-13a).Which numbered points on the screen bestcorrespond to which phasor diagram?

(a) Figure 1

(b) Figure 2

(c) Figure 3

(d) Figure 4

How much faster, in meters per second, does light travel in sapphire than in diamond? See Table 33-1.

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