Physics for Scientists and Engineers with Modern Physics
Physics for Scientists and Engineers with Modern Physics
10th Edition
ISBN: 9781337553292
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 36, Problem 16P

Show that the distribution of intensity in a double-slit pattern is given by Equation 36.9. Begin by assuming that the total magnitude of the electric field at point P on the screen in Figure 36.4 is the superposition of two waves, with electric field magnitudes

E 1 = E 0 sin ω t E 2 = E 0 sin ( ω t + ϕ )

The phase angle in ϕ in E2 is due to the extra path length traveled by the lower beam in Figure 36.4. Recall from Equation 33.27 that the intensity of light is proportional to the square of the amplitude of the electric field. In addition, the apparent intensity of the pattern is the time-averaged intensity of the electromagnetic wave. You will need to evaluate the integral of the square of the sine function over one period. Refer to Figure 32.5 for an easy way to perform this evaluation. You will also need the trigonometric identity

sin A + sin B = 2 sin ( A + B 2 ) cos ( A B 2 )

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1. A long silver rod of radius 3.5 cm has a charge of -3.9 ис on its surface. Here ŕ is a unit vector ст directed perpendicularly away from the axis of the rod as shown in the figure. (a) Find the electric field at a point 5 cm from the center of the rod (an outside point). E = N C (b) Find the electric field at a point 1.8 cm from the center of the rod (an inside point) E=0 Think & Prepare N C 1. Is there a symmetry in the charge distribution? What kind of symmetry? 2. The problem gives the charge per unit length 1. How do you figure out the surface charge density σ from a?
1. Determine the electric flux through each surface whose cross-section is shown below. 55 S₂ -29 S5 SA S3 + 9 Enter your answer in terms of q and ε Φ (a) s₁ (b) s₂ = -29 (C) Φ զ Ερ (d) SA = (e) $5 (f) Sa $6 = II ✓ -29 S6 +39
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Physics for Scientists and Engineers with Modern Physics

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