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Figure 35-27a shows the cross-section of a vertical thin film whose width increases downward because gravitation causes slumping. Figure 35-27b is a face-on view of the film, showing four bright (red) interference fringes that result when the film is illuminated with a perpendicular beam of red light. Points in the cross section corresponding to the bright fringes are labeled. In terms of the wavelength of the light inside the film, what is the difference in film thickness between (a) points a and b and (b) points b and d?

Short Answer

Expert verified

(a) The difference in the thickness between points a and b isλ2n.

(b) The difference in the thickness between points b and d isλn.

Step by step solution

01

Write the given data from the question:

  • The width of the thin film increases downward.
  • The film is illuminated with a beam of red light.
02

Determine the formulas to calculate the difference in the film thickness:

The expression to calculate the wavelength inside the film is given as follows.

L=mλ2n

Here, n is the refractive index of the medium, λis the wavelength, L is the thickness of the firm, and m is the order of the interference and it can take values 0,1,2,3........

03

(a) Calculate the difference in the film thickness of points a  and b: 

The bright fringes in the thin films produce when 2L is equal to multiple of wavelength.

2L=mλt

Substituteλnforλtinto above equation.

2L=mλnL=mλ2n

............(i)

Between points, a and b, the order of the interference is 1. Therefore,

m=1

Substitute 1 for m into equation (i).

L=1λ2nL=λ2n

Hence, the difference in the thickness between points a and b isλ2a.

04

(b) Calculate the difference in the film thickness of points b and d.

Since between points b and d, the order of the interference is 2. Therefore,

m=2

Substitute 2 for m into equation (i)

L=2λ2nL=λn

Hence, the difference in the thickness between points b and d isλn.

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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=(2.00μ³Õ/m)sin[1.26×1015t]E2=(2.00μ³Õ/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?

In a double-slit experiment, the distance between slits is5.0mm and the slits are 1.0m from the screen. Two interference patterns can be seen on the screen: one due to light of wavelength 480nm, and the other due to light of wavelength 600nm. What is the separation on the screen between the third-order (m=3) bright fringes of the two interference patterns?

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?

Figure 35-22 shows two light rays that are initially exactly in phase and that reflect from several glass surfaces. Neglect the slight slant in the path of the light inthe second arrangement.

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

In wavelengthsλ,

(b) what should that path length difference equal if the rays are to be exactly out of phase when they emerge, and

(c) what is the smallest value of that will allow that final phase difference?

In Fig. 35-45, two microscope slides touch at one end and are separated at the other end. When light of wavelength 500 nm shines vertically down on the slides, an overhead observer sees an interference pattern on the slides with the dark fringes separated by 1.2 mm. What is the angle between the slides?

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