Homework ## Chapter

25: Wave Optics

Ch 25: 5, 17, 28, 40, 44, 62, 63

25.5 Two slits 0.03 mm apart are illuminated by parallel rays and the 5th-order maximum is 14 cm from the central bright maximum on a screen 2.0 m from the slits. What is the wavelength of the light falling on the slits?Maxima (or bright fringes) occur for

d sin = m m in an integer. We can solve this for the wavelength ,

= (d sin ) / m We can solve for or sin from

tan = ^{opp}/_{adj}= 14 cm/200 cm = 0.07= 4.00

^{o}sin = sin 4.00

^{o}= 0.07d = 0.03 mm = 3 x 10

^{ - 5}mTherefore,

= [(3 x 10 ^{ - 5}m)(0.07)] / 5= 4.2 x 10

^{ - 7}m = 420 x 10^{ - 9}m = 420 nm

25.17Redlight (= 640 nm) passes through a diffraction grating and onto a screen 2.0 meters away. There the first-order maximum or bright line appears 25 cm on either side of the central maximum. How many lines per centimeter are there on the grating?First, determine the angle ,

tan = 25 cm/200 cm = 0.125 = 7.125°

We expect to find maxima or bright regions for

d sin = m Here we have only first-order maxima, with m = 1, on either side of the central maximum (m = 0) so we can solve for d, the separation distance between the "slits" of the diffraction grating,

d = m / sin = / sin d = 640 nm / sin 7.125

^{o}d = 640 nm / 0.124

d = 5160 nm = 5,160 x 10

^{ - 9}m = 5.16 x 10^{ - 6}md = 5.16 x 10

^{ - 4}cmThis is the slit separation distance. From this we can determine N the number of slits there are in a centimeter. If there were ten slits per centimeter (N = 10/cm) then the separation distance would be 0.1 cm (d = 0.1 cm). N and d are related by

N = 1/d N = 1.94 x 10

^{3}lines/cm = 1,940 lines/cm

25.28 When 634-nm light passes through a slit 0.1 mm wide and forms a diffraction pattern on a screen 1.25 m away, what is the distance from the central maximum to the center of the first minimum?For a single-slit diffraction pattern, the first minimum occurs for

D sin = or

sin = /D sin = 634 nm/0.1mm

sin = (634 x 10

^{ - 9}m)/(0.1x 10^{ - 3}m)sin = 6.34 x 10

^{ - 3}= 0.00634The distance y is given by

tan = y/L where L is the distance from the slit to the screen, L = 1.25 m

y = L tan For such small angles

tan sin y = L sin

y = (1.25 m)(0.00634)

y = 7.93 mm

25.40 A high precision telescope advertises that it can resolve the eyes of a bird (assume them 1 cm apart) at 1.5 km. What is the minimum objective diameter?The Rayleigh criterion for rsolution of two point sources is

= 1.22 / D D = 1.22 /

tan = 1.0 cm / 1.5 km = 1.0 x 10

^{ - 2}m / 1.5 x 10^{3}mtan = 6.67 x 10

^{ - 6}= 6.67 x 10

^{ - 6}radians ( = 3.8 x 10^{ - 4}^{o})Since no particular wavelength is given for the light, use a wavelength near the middle of the visible spectrum such as

550 nm = 550 x 10 ^{ - 9}mD = 1.22 /

D = 1.22 (550 x 10

^{ - 9}m) / 6.67 x 10^{ - 9}D = 10 cm

25.44 Magnesium chloride (MgCl_{2}), which has an index of refraction of 1.25, is often used for the non-reflecting coating on photographic lenses. How thick must a film of it be to have total destructive interference for 560 nm light?For a thin film like this, with n

_{1}< n_{2}and n_{2}< n_{3}, total destructive interference occurs for2 t = [ m + (1/2)] ' = [ m + (1/2)] / n m is an integer. For the thinest film, m = 0.

2 t = (1/2)(560 nm) / 1.25 t = (1/2)(560 nm) / [(1.25)(2)]

t = 112 nm

t = 1.12 x 10

^{ - 7}m

25.62 When 580-nm light passes through a 1500 line/cm diffraction grating, how many orders of maxima do you see on a screen that is 1.5 m wide and1.2 m from the grating?Recalling, from Problem 25.28, that d = 1/N, we can determine the slit separation distance d from N the number of lines/cm,

d =6.67 x 10 ^{ - 6}mMaxima for diffraction patterns occur for

d sin = m The largest possible angle

_{max}istan _{max}= 1.5 m / 1.2 m = 1.25

_{max}= 51.3°sin

_{max}= 0.78Then the largest possible order mmax is given by

m = d sin /

m

_{max}= d sin_{max}/m

_{max}= (6.67 x 10^{ - 6}m)(0.78)/(580 x 10^{ - 9}m)m

_{max}= 8.97But m must be an integer! What does m

_{max}= 8.97 mean? The maximum order of diffraction pattern (bright area) that we will see will bem = 8 The next bright diffraction pattern, for m = 9, will be just off the edge of the screen.

25.63 An observer looks at a light source through a grating having 1000 lines/cm. If= 500 nm, how many images of the source can be seen? (Hint: For what order is q = 90°?)Of course, this is similar to the previous problem. We can determine the slit separation distance d from N the number of lines/cm,

d =1.0 x 10 ^{ - 5}mMaxima for diffraction patterns occur for

d sin = m The largest possible angle

_{max}is 90^{o}or sin_{max}= 1.0m _{max}= d sin_{max}/m

_{max}= (1.0 x 10^{ - 5}m) (1.0)/ (500 x 10^{ - 9}m)m

_{max}= 20

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