# Quimica

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Chapter 37 Solutions

37.1

λL (632.8 × 10- 9)(5.00) ∆ybright = d = m = 1.58 cm 2.00 × 10- 4

37.2

y bright

λL = m d

For m = 1,

3.40 × 10 − 3 m 5.00 × 10 − 4 m yd λ= = = 515 nm L 3.30 m

(

)(

)

37.3

Note, with the conditions given, the small angle approximation does not work well. That is, sin θ, tan θ, and θ are significantly different. The approach to be used isoutlined below. (a) At the m = 2 maximum, 400 m tan θ = 1000 m = 0.400

θ = 21.8° so
(b)

λ=

d sin θ (300 m) sin 21.8° = = 55.7 m m 2

The next minimum encountered is the m = 2 minimum; and at that point,
1 5 d sin θ =  m + 2 λ which becomes d sin θ = 2 λ  

or so

5λ 5(55.7 m) sin θ = 2d = 2(300 m) = 0.464 y = (1000 m) tan 27.7° = 524 m

and

θ = 27.7°

Therefore, the carmust travel an additional 124 m .

37.4 (a) (b) (c)

v 354 m/s λ = f = 2000/s = 0.177 m d sin θ = m λ d sin θ = m λ so so (0.300 m) sin θ = 1(0.177 m) d sin 36.2° = 1(0.0300 m) so and and

θ = 36.2°
d = 5.08 cm

(1.00 × 10- 6 m) sin 36.2° = 1λ f= 3.00 × 108 m/s c = = 508 THz 5.90 × 10–7 m λ

λ = 590 nm

2

Chapter 37 Solutions37.5

For the tenth minimum, m = 9. Using Equation 37.3, Also, tan θ = 9.5λ sin θ y . L

sin θ =

1 λ 9+  d 2

For small θ , sin θ ≈ tan θ . Thus,

d=

=

9.5λ L 9.5(589 × 10–9 m)(2.00 m) = y 7.26 × 10–3 m

= 1.54 × 10–3 m = 1.54 mm

Goal Solution Young's double-slit experiment is performed with 589-nm light and a slit-to-screen distance of 2.00 m . The tenth interferenceminimum is observed 7.26 mm from the central maximum. Determine the spacing of the slits. G: For the situation described, the observed interference pattern is very narrow, (the minima are less than 1 mm apart when the screen is 2 m away). In fact, the minima and maxima are so close together that it would probably be difficult to resolve adjacent maxima, so the pattern might look like a solid blur tothe naked eye. Since the angular spacing of the pattern is inversely proportional to the slit width, we should expect that for this narrow pattern, the space between the slits will be larger than the typical fraction of a millimeter, and certainly much greater than the wavelength of the light (d >> λ = 589 nm). Since we are given the location of the tenth minimum for this interference pattern, weshould use the equation for destructive interference from a double slit. The figure for Problem 7 shows the critical variables for this problem. d sin θ = m + sin θ =

O:

A:

In the equation The first minimum is described by m = 0 and the tenth by m = 9 : Also, tan θ = y L , but for small θ , sin θ ≈ tan θ . Thus, 9.5(5890·10 -10 m)(2.00 m) = 1.54·10 -3 m = 1.54 mm = 1.54 mm 7.26·10 -3 m

(1 2 1 2

)λ ,

λ 9+ d

( )

d=

9.5λ 9.5λ L = y sin θ

d=

L:

The spacing between the slits is relatively large, as we expected (about 3 000 times greater than the wavelength of the light). In order to more clearly distinguish between maxima and minima, the pattern could be expanded by increasing the distance to the screen. However, as L is increased, the overall pattern wouldbe less bright as the light expands over a larger area, so that beyond some distance, the light would be too dim to see.

Chapter 37 Solutions

3

*37.6

340 m/s λ = 2000 Hz = 0.170 m Maxima are at m=0 m=1 m=2 m=3 gives gives gives gives d sin θ = m λ :

θ = 0° λ 0.170 m sin θ = d = 0.350 m 2λ sin θ = d = 0.971 sin θ = 1.46 θ = 29.1° θ = 76.3°
No solution.

1 Minima at d sin θ =  m+ 2  λ:  

m=0 m=1 m=2

λ sin θ = 2d = 0.243 3λ gives sin θ = 2d = 0.729 gives No solution.
gives

θ = 14.1° θ = 46.8°

So we have maxima at 0°, 29.1°, and 76.3° and minima at 14.1° and 46.8°.

37.7

(a)

For the bright fringe, y bright = y= mλ L d where m=1

(546.1 × 10 −9 m)(1.20 m) = 2.62 × 10 −3 m = 2.62 mm 0.250 × 10 −3 m

(b)

For the dark bands, ydark = y2 − y1 =...