Figure 35-25 shows two sources s1 and s2 that emit radio waves of wavelengthλin all directions. The sources are exactly in phase and are separated by a distance equal to 1.5λ . The vertical broken line is the perpendicular bisector of the distance between the sources.

(a) If we start at the indicated start point and travel along path 1, does the interference produce a maximum all along the path, a minimum all along the path, or alternating maxima and minima? Repeat for

(b) path 2 (along an axis through the sources) and

(c) path 3 (along a perpendicular to that axis).

Short Answer

Expert verified

(a) The interference produces maxima all along path 1.

(b) The interference produces minima all along path 2.

(c) The interference produces alternating maxima and minima all along path 3.

Step by step solution

01

Given data

Distance between the two sources = 1.5λ

02

Interference fringe path difference

The path difference of two rays creating a bright fringe of order m for slit separation d, screen distance D, and wavelength λis

ΔL=mλ …(i)

The path difference of two rays creating a dark fringe of order m for slit separation d, screen distance D and wavelength λis

ΔL=(m+12)λ…(ii)

03

(a) Determining the path difference between light from the two sources in path 1

For any point y path 1, the path difference between light rays from the two sources is

ΔL=0.75λ2+y2-0.75λ2+y2=0

This is equal to equation (i) with m=0. Thus path 1 has bright fringes all along.

04

(b) Determining the path difference between light from the two sources in path 2

For any point on path 2 at a distance a from source 2, the path difference between light rays from the two sources is

ΔL=a+1.5λ-a=1.5λ

This is equal to equation (ii) with m=1. Thus path 1 has dark fringes all along.

05

(c) Determining the path difference between light from the two sources in path 3 

For any point on path 3 at a horizontal distance a from source 2 and vertical distance y, the path difference between light rays from the two sources is

ΔL=a+1.5λ2+y2-a2+y2

This function has a maximaΔL=1.5λat y=0 and tends to 0 asy. Thus there is a dark fringe at y=0, a bright fringe when ΔLreduces to 1λ, another dark fringe when ΔLreduces to 0.5λand finally a bright fringe as y. Thus there are alternating bright and dark fringes.

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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?

Figure 35-40 shows two isotropic point sources of light (S1and S2) that emit in phase at wavelength 400 nm and at the same amplitude. A detection point P is shown on an x-axis that extends through source S1. The phase difference ϕbetween the light arriving at point P from the two sources is to be measured as P is moved along the x axis from x=0 out to x=+.The results out to xs=10×10-7m are given in Fig. 35-41. On the way out to + , what is the greatest value of x at which the light arriving at from S1is exactly out of phase with the light arriving at P from S2?

Figure 35-26 shows two rays of light, of wavelength 600nm, that reflectfrom glass surfaces separated by 150nm. The rays are initially in phase.

(a) What is the path length difference of the rays?

(b) When they have cleared the reflection region, are the rays exactly in phase, exactly out of phase, or in some intermediate state?

Figure 35-46a shows a lens with radius of curvature lying on a flat glass plate and illuminated from above by light with wavelength l. Figure 35-46b (a photograph taken from above the lens) shows that circular interference fringes (known as Newton’s rings) appear, associated with the variable thickness d of the air film between the lens and the plate. Find the radii r of the interference maxima assumingr/R1.

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 Lin nanometres, and the wavelength λin nanometres 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.

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