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Question: 

Suppose a thin piece of glass is placed in front of the lower slit in Fig. 24–7 so that the two waves enter the slits $180 ^\circ$ out of phase (Fig. 24–58). Draw in detail the interference pattern seen on the screen.

Problem 11.

Figure 24-58.

How the wave theory explains the pattern of lines seen in the double-slit experiment.

Figure 24-7 How the wave theory explains the pattern of lines seen in the double-slit experiment. (a) At the center of the screen, waves from each slit travel the same distance and are in phase. [Assume $l \gg d$.]

How the wave theory explains the pattern of lines seen in the double-slit experiment.

Figure 24-7 How the wave theory explains the pattern of lines seen in the double-slit experiment. (b) At this angle $\theta$, the lower wave travels an extra distance of one whole wavelength, and the waves are in phase; note from the shaded triangle that the path difference equals $\textrm{d sin} \; \theta$.

How the wave theory explains the pattern of lines seen in the double-slit experiment.

Figure 24-7 How the wave theory explains the pattern of lines seen in the double-slit experiment. (c) For this angle $\theta$, the lower wave travels an extra distance equal to one-half wavelength, so the two waves arrive at the screen fully out of phase.

How the wave theory explains the pattern of lines seen in the double-slit experiment.

Figure 24-7 How the wave theory explains the pattern of lines seen in the double-slit experiment. (d) A more detailed diagram showing the geometry for parts (b) and (c).

Source: Giancoli, Douglas C., Physics: Principles with Applications, 7th Edition, 2014.

Quick Answer: 

constructive: $d \sin \theta = \lambda \left( m - \dfrac{1}{2} \right), m=1,2,3,...$
destructive: $d \sin \theta = m \lambda, m = 0,1,2,3,...$
The interference pattern is reversed.

Giancoli 7th Edition, Chapter 24, Problem 11

(3:19)

Chapter 24, Problem 11 is solved.

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