Physics Solution Manual for 1100 and 2101

# Repeating the calculation for the second order

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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.

Ask a homework question - tutors are online