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In Fig. 35-31, a light wave along ray r1reflects once from a mirror and a light wave along ray r2reflects twice from that same mirror and once from a tiny mirror at distance Lfrom the bigger mirror. (Neglect the slight tilt of the rays.) The waves have wavelength 620 nm and are initially in phase. (a) What is the smallest value of Lthat puts the final light waves exactly out of phase? (b) With the tiny mirror initially at that value of L, how far must it be moved away from the bigger mirror to again put the final waves out of phase?

Short Answer

Expert verified
  1. The smallest value of that puts the final light waves exactly out of phase is 155nm.
  2. The tiny mirror should be moved 310nmaway from the bigger mirror to again put the final waves out of phase.

Step by step solution

01

Given information

The wavelength of both light rays is,λ=620nm

02

Destructive interference

When two waves of equal frequency and opposite phase meet each other then they cancel each other out and results in the ‘destructive interference’.

For destructive interference, the value of the path difference between two waves is given by,

Δx=2n+1λ2

Here,λ is the wavelength of both waves and n=1,2,3,...is the number of fringes.

03

(a) Smallest value of distance between tiny and bigger mirror

According to the question, the waveW2reflects twice from that same mirror and once from a tiny mirror located at a distance L. Has no substantive effect on the calculations,

The value of the phase difference after two reflections will be,

2λ2=λ

The distance travelled by wave W2is 2Lgreater than the distance travelled by wave W1.

Then the smallest value of Lthat puts the wave role="math" localid="1663049389226" W2a half-wavelength "behind" wave W1is given by,

2L=λ2L=λ4L=620nm4L=155nm

Hence, the smallest value ofL that puts the final light waves exactly out of phase is155nm .

04

(b) Distance moved by tiny mirror

The fact that wave W2reflects two additional times has no substantive effect on the calculations. Since two reflections amount to a 2λ2=λphase difference, which is effectively not a phase difference at all. The substantive difference between W2and W1is the extra distance 2L travelled by W2. Destructive interference will again appear ifW2is3λ2"behind" the other wave.

In this case, 2L'=3λ2and the difference is,

L'-L=3λ4-λ4L'-L=λ2L'-L=620nm2L'-L=310nm

Hence, the tiny mirror should be moved 310nmaway from the bigger mirror to again put the final waves out of phase.

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Most popular questions from this chapter

A 600nm-thick soap film n=1.40in air is illuminated with white light in a direction perpendicular to the film. For how many different wavelengths in the 300to 700nm range is there (a) fully constructive interference and (b) fully destructive interference in the reflected light?

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, n2, and 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 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.

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, n2andn3, 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.

In Fig. 35-4, assume that two waves of light in air, of wavelength 400nm, are initially in phase. One travels through a glass layer of index of refraction n1=1.60and thickness L. The other travels through an equally thick plastic layer of index of refraction n2=1.50. (a) What is the smallest value Lshould have if the waves are to end up with a phase difference of 5.65 rad? (b) If the waves arrive at some common point with the same amplitude, is their interference fully constructive, fully destructive, intermediate but closer to fully constructive, or intermediate but closer to fully destructive?

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 rayr3(the light does not reflect inside material 2) andr4(the light reflects twice inside material 2). The waves ofr3and 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 n3the type of interference, the thin-layer thickness Lin nanometers, and the wavelength λin nanometers of the light as measured in air. Whereλis 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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