Physics Solution Manual for 1100 and 2101

Repeating the calculation for the second order

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Unformatted text preview: milar to that outlined in Example 6 in the text. According to Equation 27.4, the dark fringes for single slit diffraction are located by sin θ = mλ/W. For first-order fringes, m = 1, and sin θ 1 = λ / W . For second-order fringes, m = 2, and sin θ 2 = 2 λ / W . Therefore, we find that θ 1 = sin –1 480 λ F I = sin F × 10 m I = 1.4 ° G J G.0 × 10 m J HK H W 2 K O 2 2 Fλ I = sin Lc × 10 mh= 2.8° G J M2480× 10 m P H K M .0 W P N Q –9 –1 –5 –9 θ 2 = sin –1 –1 –5 From Example 6 we know that the distance y is given by y = L tan θ, where L is the distance between the slit and the screen. Therefore, it follows that c y 2 – y 1 = L tan θ 2 – tan θ 1 0 b h= b.50 mgtan 2.8° – tan 1.4° g= 0 . 012 m 30. REASONING The distance y between the center of the diffraction pattern and a particular dark fringe depends upon the distance L between the slit and the screen and the angle θ at which the dark fringe appears y = L tan θ (1) For light with a wavelength λ passing through a slit of width W, the angles of the dark fringes are given by sin θ = m λ (Equation 27.4), where m is an integer (m = 1, 2, 3, …). W The screen will only be completely dark when both diffraction patterns have a dark fringe at the same angle θ . We will use Equation 27.4 to determine the integers (m1 and m2) for which the two wavelengths (λ1 = 632 nm and λ2 = 474 nm) have dark fringes at the same angle. Then, Equation (1) will determine the distance y. SOLUTION For an angle θ at which both diffraction patterns have dark fringes, Equation 27.4 gives sin θ = m1 λ1 W and sin θ = m2 Setting the right sides of Equations (2) equal, we see that λ2 W (2) Chapter 27 Problems m1 λ1 W = m2 λ2 m1 λ2 474 nm == = 0.75 m2 λ1 632 nm or W 1453 (3) The first dark fringes of the two diffraction patterns do not coincide, because setting m1 = m2 = 1 yields a ratio of m1/m2 = 1/1 = 1, which does not satisfy Equation (3). But we can see that other dark fringes do coincide, because Equation (3) is satisfied when m1 = 3 and m2 = 4 (m1/m2 = 3/4 = 0.75), or when m1 = 6 and m2 = 8 (m1/m2 = 6/8 = 0.75), and so forth. The first time the dark fringes overlap occurs when m1 = 3 and m2 = 4. Solving the first of Equations (2) for θ , and taking m1 = 3 yields θ = sin −1 m1 632 ×10−9 m o = sin −1 3 7.15 ×10−5 m = 1.52 W λ1 Then, from Equation (1), we have that y = L tan θ = (1.20 m ) tan1.52o = 3.18 ×10−2 m = 3.18 cm 31. SSM REASONING The width of the central bright fringe is defined by the location of the first dark fringe on either side of it. According to Equation 27.4, the angle θ locating the first dark fringe can be obtained from sin θ = λ/W, where λ is the wavelength of the light and W is the width of the slit. According to the drawing, tan θ = y/L, where y is half the width of the central bright fringe and L is the distance between the slit and the screen. First dark fringe y θ W Midpoint of central bright fringe L SOLUTION Since the angle θ is small, we can use the fact that sin θ ≈ tan θ. Since sin θ = λ/W and tan θ = y/L, we have λ W = y L or W= λL y 1454 INTERFERENCE AND THE WAVE NATURE OF LIGHT Applying this result to both slits gives W2 W1 W2 = W1 y1 y2 c = = 3.2 × 10 −5 λ L / y 2 y1 = λ L / y1 y 2 m 1 cm g hb.2 cmg 1.9 b= 1 2 1 2 2 .0 × 10 −5 m 32. REASONING AND SOLUTION It is given that 2y = 450W and L = 18 000W. We know λ/W = sin θ. Now sin θ ≈ tan θ = y/L, so λ W = 225 W y = = L 18 000 W 0.013 33. REASONING AND SOLUTION The minimum angular separation of the cars must be θmin = 1.22 λ/D, and the separation of the cars is y = Lθmin = 1.22 Lλ /D. a. For red light, λ = 665 nm, and y = (1.22)(8690 m)(665 × 10 m)/(2.00 × 10 m) = 3.53 m –9 –3 b. For violet light, λ = 405 nm, and y = (1.22)(8690 m)(405 × 10–9 m)/(2.00 × 10–3 m) = 2 .15 m 34. REASONING The drawing shows the two stars, separated by a distance s, that are a distance r from the earth. The angle θ (in radians) is defined as the arc length divided by the radius r (see Equation 8.1). Assuming that r >> s, the arc length is very nearly equal to the distance s, so we can write θ = s/r. Since s is known, this relation will allow us to find the distance r if we can determine the angle. If the stars can just be seen as separate objects, we know that the minimum angle θmin between them is given by Equation 27.6 as θ min ≈ 1.22λ / D , where λ is the wavelength of the light being observed and D is the diameter of the telescope’s objective. By combining these two relations, we will be able to find the distance r. s r θ Earth Chapter 27 Problems SOLUTION Solving the relation θ θ = θmin ≈ 1.22λ / D , we obtain r= s θ s = 1.22 λ D = 1455 = s/r for r, and substituting in the relation 3.7 ×1011 m = 5.6 × 1017 m 550 ×10−9 m 1.22 1.02 m 35. SSM REASONING According to Rayleigh's criterion, the two taillights must be separated by a distance s sufficient to subtend an angle θ min ≈ 1.22 λ / D at the pupil of the observer's eye. Recalling that this angle must be expressed in ra...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.

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