Home Work 17

Home Work 17

13-1 Manufacturers of wire (and other objects of small dimension) sometimes use a laser to continually monitor the thickness of the product. The wire intercepts the laser beam, producing a diffraction pattern like that of a single slit of the same width as the wire

diameter (Fig. 36-39). Suppose a helium–neon laser, of wavelength 632.8 nm, illuminates a wire, and the diffraction pattern appears on

a screen at distance L = 2.60 m. If the desired wire diameter is 1.37 mm, what is the observed distance between the two tenth-order minima (one on each side of the central maximum)?

Sol:

From y = mL/a we get

(632.8 nm)(2.60)

[10 ( 10)] 24.0 mm . 1.37 mm

m L L

y m

a a

 

 

        

 

17-2 Floaters. The floaters you see when viewing a bright, featureless background are diffraction patterns of defects in the vitreous humor that fills most of your eye. Sighting through a pinhole sharpens the diffraction pattern. If you also view a

small circular dot, you can approximate the defect’s size. Assume that the defect diffracts light as a circular aperture does. Adjust the dot’s distance L from your eye (or eye lens) until the dot and the circle of the first minimum in the diffraction pattern appear to have the same size in your view. That is, until they have the same diameter D’ on the retina at distance L’ = 2.0 cm from the front of the eye, as suggested in Fig. 36-44a, where the angles on the two sides of the eye lens are equal. Assume that the wavelength of visible light is λ= 550 nm. If the dot has diameter D

= 2.0 mm and is distance L = 45.0 cm from the eye and the defect is x = 6.0 mm in front of the retina (Fig. 36-44b), what is the diameter of the defect?

Sol:

From Fig. 36-44(a), we find the diameter D on the retina to be 2.00 cm

(2.00 mm) 0.0889 mm 45.0 cm

D DL L

    .

Next, using Fig. 36-44(b), the angle from the axis is

1 / 2 1 0.0889 mm / 2

tan tan 0.424

6.00 mm D

  ^{ } ^x ^ ^{ } ^ 

    .

Since the angle corresponds to the first minimum in the diffraction pattern, we have sin1.22 / d , where  is the wavelength and d is the diameter of the defect. With

550 nm,

 we obtain

1.22 1.22(550 nm) 5

9.06 10 m 91 m sin sin(0.424 )

d  



     

 .

17-3 Figure 36-49 is a graph of intensity versus angular position θ for the diffraction of an x-ray beam by a crystal. The horizontal scale is set byθ_s = 2.00°. The beam consists of two wavelengths, and the spacing between the reflecting planes is 0. 94 nm. What are the (a) shorter and (b) longer wavelengths in the beam?

Sol:

We use Eq. 36-34.

(a) From the peak on the left at angle 0.75° (estimated from Fig. 36-49), we have

_2d sin₁2 0 94

b

g b g

This is the shorter wavelength of the beam. Notice that the estimation should be viewed as reliable to within ±2 pm.

(b) We now consider the next peak:

_2d sin₂ 2 0 94

b

g

This is the longer wavelength of the beam. One can check that the third peak from the left is the second-order one for 1.

17-4 In Fig. 36-50, first-order reflection from the reflection planes shown occurs when an x-ray beam of wavelength 0.260 nm makes an angleθ= 63.8°

with the top face of the crystal. What is the unit cell size a₀? Sol:

The angle of incidence on the reflection planes is  = 63.8° – 45.0° = 18.8°, and the plane-plane separation is d a₀ 2 . Thus, using 2d sin  = , we get

a₀ 2d 2 0 260

2 18 8 0 570

  

 



sin

.

sin . .



nm nm.

17-5 A diffraction grating having 180 lines/mm is illuminated with a light signal containing only two wavelengths, λ1 = 400 nm and λ2 = 500 nm. The signal is incident perpendicularly on the grating. (a) What is the angular separation between the second-order maxima of these two wavelengths? (b) What is the smallest angle at which two of the resulting maxima are

superimposed? (c) What is the highest order for which maxima for both wavelengths are present in the diffraction pattern?

Sol

(a) Since d = (1.00 mm)/180 = 0.0056 mm, we write Eq. 36-25 as

1 1

sin m sin (180)(2)

 ^{ } d^ ^ 

where _ 4 10^4mm and _  5 10^4mm. Thus,  ₂₁2 1. . 

(b) Use of Eq. 36-25 for each wavelength leads to the condition

m₁₁m₂₂

for which the smallest possible choices are m₁ = 5 and m₂ = 4. Returning to Eq. 36-25, then, we find

 

4

1 1 1 1 10 mm) 1

sin sin sin 0.36 21 .

0.0056 mm m

 ^{ } d^{ } ^{ }^{  } ^  

(c) There are no refraction angles greater than 90°, so we can solve for “m_max” (realizing it might not be an integer):

max 4

2 2

sin 90 0.0056 mm 10 mm 11

d d

m     _ 

  

where we have rounded down. There are no values of m (for light of wavelength 2) greater than m = 11.

17-6 The D line in the spectrum of sodium is a doublet with wavelengths 589.0 and 589.6 nm.

Calculate the minimum number of lines needed in a grating that will resolve this doublet in the second-order spectrum.

Sol

Letting R = / = Nm, we solve for N:

 

 

589.6 nm 589.0 nm / 2 2 589.6 nm 589.0 nm 491.

N m





   

 

17-7 With a particular grating the sodium doublet (589.00 nm and 589.59 nm) is viewed in the third order at 10° to the normal and is barely resolved. Find (a) the grating spacing and (b) the total width of the rulings.

Sol

(a) From d sin  = m we find

d m

 

  

_avg sin

nm nm = 10 m.

 ^{3 589 3} 

10 10 10⁴

.

sin .

b g

(b) The total width of the ruling is

L Nd R

m d d

 

F

H GIK J

nm nm m = 3.3 mm.



589 3 10

3 589 59. 589 00 3 3 10³

. . .

b gb g

b

g