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 = 4.00o
sin = sin 4.00o = 0.07
d = 0.03 mm = 3 x 10 - 5 m
Therefore,
= [(3 x 10 - 5 m)(0.07)] / 5 = 4.2 x 10 - 7 m = 420 x 10 - 9 m = 420 nm
25.17 Red light ( = 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 7.125o
d = 640 nm / 0.124
d = 5160 nm = 5,160 x 10 - 9 m = 5.16 x 10 - 6 m
d = 5.16 x 10 - 4 cm
This 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 103 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 = 0.00634
The 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 103 m
tan = 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 m D = 1.22 /
D = 1.22 (550 x 10 - 9 m) / 6.67 x 10 - 9
D = 10 cm
25.44 Magnesium chloride (MgCl2), 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 n1< n2 and n2 < n3, total destructive interference occurs for
2 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 m Maxima for diffraction patterns occur for
d sin = m The largest possible angle max is
tan max = 1.5 m / 1.2 m = 1.25 max = 51.3°
sin max = 0.78
Then the largest possible order mmax is given by
m = d sin /
mmax = d sin max /
mmax = (6.67 x 10 - 6 m)(0.78)/(580 x 10 - 9 m)
mmax = 8.97
But m must be an integer! What does mmax = 8.97 mean? The maximum order of diffraction pattern (bright area) that we will see will be
m = 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 m Maxima for diffraction patterns occur for
d sin = m The largest possible angle max is 90o or sin max = 1.0
mmax= d sin max / mmax = (1.0 x 10 - 5 m) (1.0)/ (500 x 10 - 9 m)
mmax = 20
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