37 CHAPTER INTERFERENCE AND DIFFRACTION
.
-
--
-
Section 37-2:
Double-Slit Interference
Section 37-3: Multiple-Slit Interference and Diffraction Gratings Problem 13. In a 5-slit system, how many minima lie betwee* the zeroth-order and first-order maxima? Solution In an N-slit system with slit separation d (illuminated by normally incident plane waves), the main maxima occur for angles sin 0 = mX/d, and minima for sin 8 = mJX/Nd (excluding m' equal to zero or multiples of N ). Between two adjacent maxima, say m' = m N and (m 1)N, there are N - 1 minima. (The number of integers between m N and (m + l ) N is (m l ) N - m N - 1 = N - 1, because the limits are not included.) For N = 5, the number of minima is 4.
Problem 3. A double-slit experiment has slit spacing 0.12 mm. (a) What should be the slit-tc-screen distance L if the bright fringes are to be 5.0 mm apart when the slits are illuminated with 633-nm laser light? (b) What will be the fringe spacing with 480-nm light? Solution The particular geometry of this type of double-slit experiment is described in the paragraphs preceding Equations 37-2a and b. (a) The spacing of bright fringes on the screen is Ay = XL/d, so L = (0.12 m_ _- m)/(633 nm] = 94.8 cm. ( b) For two m)(5 m different wavelengths, the ratio of the spacings is a y f / A y = Af/A; therefore Ay' = (5 mm)(480/633) = 3.79 mm.
+
+
Problem 17. Green light at 520 nm is diffracted by a grating with 3000 lines per cm. Through what angle is the light diffracted in (a) first and (b) fifth order? Solution For light normally incident on a diffraction grating, maxima occur a t angles 6' = sin-'(mX/d), where d is the grating spacing (equal to the reciprocal of the number of lines per meter), and m is the order. (a) In first order, 6' = sin-'(520 nm x 3000/cm) = 8.97", and (b) in fifth order, O5 = s inq1(5 sin 8.97') = 51.3'. Problem 23. Estimate the number of lines per cm in the grating used to produce Fig. 37-15. Solution The number of lines per cm (lid in cm-') is easily estimated from the angular position of the central 550-nm line in a particular order, as shown in the figure; that is, l /d = sin 6'lmA. For example, in fifth order, this line is at 6' = 61' (average of right and left values), so l /d = sin61/5(550 nm) = 3.18x103/cm or a bout 3200 lines/cm.
Problem 7. Light shines on a pair of slits whose spacing is three times the wavelength. Find the locations of the first- and second-order bright fringes on a screen 50 cm from the slits. Hint: Do Equations 37-2 apply7 Solution Since d = 3X, t he angles are not small, and Equations 37-2 do not apply. The interference maxima occur a t angles given by Equation 37-la, 6' = sin-](mX/d) = s ill-'(m/3), so only two orders are present, for values of m = 1 and 2 (6' < 90'). I f we assume that the slit/screen geometry is as shown in Fig. 37-6, then y = L tan 6' = L tan(sin-' (77213)) = ~m/ (Consider a right triangle with hypotenuse of 3 and opposite side m, or use tan 6' = sin 6'1 d m . )For m = 1 and 2, and L = 50 cm, this gives yl = (50 c m)(l/&) = 17.7 cm, and yz = (50 c m ) ( 2 / 4 ) = 44.7 cm.
Jw.
--
-
Problem 9. For a double-slit experiment with slit spacing 0.25 mm and wavelength 600 nm, at what angular position is the path difference equal to one-fourth of the wavelength? Solution If we set the path difference equal to a quarter wavelength, we obtain d sin 6' = X/4, or 6' w sin 6' = 600 nm/4(0.25 mm) = 6 x 1 0 ~ad 0.0344". r~
I
-
-qoO
I, -40.
I
- 30.
1
.I
8
0
3. 0
dob
90.
FIGURE 37-15 Problem 23 Solution.
Problem 29. You wish to resolve the calcium-H line at 396.85 nm from the hydrogen-&line at 397.05 nm in a first-order spectrum. To the nearest hundred, how many lines should your grating have? Solution From Equation 37-6, N = X/m AX = 397/(1 x0.2) = 1985 2000 lines. (This is the number of lines that must be illuminated.) Problem 30. X-ray diffraction in potassium chloride (KCI) results in a first-order maximum when the x rays graze the crystal plane at 8.5'. If the x-ray wavelength is 97 pm, what is the spacing between crystal planes? Solution From the Bragg condition (Equation 37-7), one finds d = mX/2 sin B = l (97 pm)/2 sin 8.5" = 328 pm = 3.28 A. Section 37-4: T hin Films and Interferometers
Problem 41. Two perfectly flat plates glass are separated at one end by a piece of paper 0.065 mm thick. A source of 550-nm light illuminates the plates from above, as shown in Fig. 37-43. How many bright bands appear to an observer looking down on the plates?
/
Incident light
FIGURE 37-43 Problems 41, 42, 43, 72.
Problem 35. As a soap bubble (n = 1.33) evaporates and thins, the reflected colors gradually disappear. (a) What is its thickness just as t he last vestige of color vanishes? (b) What is the last color seen? Solution T he minimum thickness of the bubble, which produces interference colors, is dmin = Xmin/4n,where A min is the shortest visible wavelength, normally 400 nm violet light. (See the solution to Problem 31.) Thus, d,i, = 400 nm/4(1.33) = 75.2 nm. Problem 39. An oil film with refractive index 1.25 floats on water. The film thickness varies from 0.80 pm t o 2.1 pm. If 630-nm light is incident normally on the film, a t how many locations will it undergo enhanced reflection? Solution I n a thin film of oil between air and water (nair < n,, < nwt..), there are 180' phase changes for reflection a t both boundaries. Therefore, for norm all^ incident light, Equation 37-8b gives the condition for constructive interference, d = mA/2n = m(630 nm)+ 2(1.25) = (0.252 p m)m. Varying thickness of 0.80 pm 5 d 1 2.1 pm implies that 3.17 5 m 5 8.33. Since m is an integer, 4 5 m 5 8, or 5 bright maxima occur a t locations corresponding to the allowed integers from 4 to 8.
Solution Equation 37-8a applies to constructive interference for normally incident light on the thin, wedge-shaped film of air between glass surfaces (although in this case, the 180" phase change affects rays reflected from the bottom surface of the film). Thus, d = ( m + 4 )x X/2nair = ( m 4)(550 nm)/2 = ( m $ )(275 n m). The thickness of the film varies between 0 and 0.065 mm, s o m varies between 0 and [(0.065 mm/275 nm) - = 235. Thus, there are 236 bright bands visible from above (since m = 0 counts as the first bright fringe).
+
+
41
Problem 45. What is the wavelength of light used in a Michelson interferometer if 550 bright fringes go by a fixed point when the mirror moves 0.150 mm? Solution * Each bright fringe shift corresponds to a path difference of one wavelength. The path changes by twice the distance moved by the mirror. Thus, 550X = 2x0.15 mm, or X = 545 nm.
-
.
Sections 37-6 and 37-7: Single-Slit Diffraction and the Diffraction Limit Problem 49. For what ratio of slit width to wavelength will the first minima of a s ingleslit diffraction pattern occur a t f90' Solution W hen 8 = 90, in Equation 37-9, and m = 1 for the first minimum, then a/X = 1.
--
-
Problem 57. A c amera has an f 11.4 lens, meaning that the r atio of focal length to lens diameter is 1.4. Find the smallest spot diameter (defined as the diameter of the first diffraction minimum) to which this lens can focus parallel light with 580-nm wavelength.
-
~-
Solution T he diffraction limit for a lens opening of diameter D, focusing light of wavelength X is Omin = 1.22 AID. T he radius of a spot, at the focal length of the lens, with t his angular spread, is T = fOmi, ( the spot radius equals the distance between the central maximum and first minimum). The minimum spot diameter is, therefore, 2f O min = 2(1.22)X f / D = 2(1.22)(550 n m) x 1.4 z! .0 pm (since f / D is the f-ratio). 2
. .
~
.
-.
.
Problem 59. While driving a t night, your eyes' irises have dilated t o 3.1-mm diameter. If your vision were diffraction-limited, what would be the greatest distance at which you could see a s distinct the two headlights of an oncoming car, which are spaced 1.5 m apart? Take X = 550 nm. Solution If we use the Rayleigh criterion (Equation 37-13b for small angles) to e stimate the diffraction-limited angular resolution of the eye, at a pupil diameter of 3.1 mm, in light of wavelength 550 nm, we obtain O min = 1.22(550 nm)/(3.1 m m) = 2 . 1 6 ~ 1 0z~ 5". r 4~ (Actually, the wavelength inside the eye is different, A' = Xln, because of the average index of refraction of the eye.) This angle corresponds to a linear separation of y = 1.5 m a t a distance of T = y/Omin= 1 .5 m / 2 . 1 6 ~ 1 0 - = 6.93 km 4 mi. Although other ~ factors determine visual acuity, this is a reasonable ballpark estimate.
-