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
,
We can solve for
tan
sin
d = 0.03 mm = 3 x 10 - 5 m
Therefore,
25.17 Red light (
First, determine the angle
tan
We expect to find maxima or bright regions for
d sin 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 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 sin
sin
sin
The distance y is given by
tan 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
D = 1.22
tan
tan
Since no particular wavelength is given for the light, use a wavelength near the middle of the visible spectrum such as
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 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 The largest possible angle
tan
sin
Then the largest possible order mmax is given by
m = d sin
mmax = d sin
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
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 The largest possible angle
mmax= d sin mmax = (1.0 x 10 - 5 m) (1.0)/ (500 x 10 - 9 m)
mmax = 20
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