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Physics 41 Chapter 38 HW Key
1. Helium–neon laser light (632..8 nm) is sent through a 0.300-mm-wide single slit. What is the width
of the central maximum on a screen 1.00 m from the slit?
sin θ=
λ
=
a
6.328 × 10−7
= 2.11× 10−3
3.00 × 10−4
y
= tan θ ≈ sin θ ≈ θ (for small θ)
1.00 m
2y = 4.22 mm
2. A beam of monochromatic green light is diffracted by a slit of width 0.550 mm. The diffraction
pattern forms on a wall 2.06 m beyond the slit. The distance between the positions of zero intensity
on both sides of the central bright fringe is 4.10 mm. Calculate the wavelength of the light.
y
λ
P38.2 The positions of the first-order minima are ≈ sin θ =
± . Thus, the spacing between these two minima is
L
a
λ
∆y =
2   L and the wavelength is
 a
−3
−3
 ∆y  a   4.10 × 10 m  0.550 × 10 m 
λ =
=
=


  
2
2.06 m
 2  L  


547 nm
Coherent light of wavelength 501.5 nm is sent through two parallel slits in a large, flat wall. Each slit is
0.700 µ m wide. Their centers are 2.80 µ m apart. The light then falls on a semicylindrical screen, with
its axis at the midline between the slits. (a) Predict the direction of each interference maximum on the
screen as an angle away from the bisector of the line joining the slits. (b) Describe the pattern of light
on the screen, specifying the number of bright fringes and the location of each. (c) Find the intensity of
light on the screen at the center of each bright fringe, expressed as a fraction of the light intensity Imax
at the center of the pattern.
Solution:
First some review:
The total intensity of the double slit interference
pattern is a combination of single slit diffraction and
the double slit interference. The cosine squared term
gives the intensity of the fringes inside the single slit
maxima. The sine squared terms is the ‘envelope’ that
shapes the fringes.
 πd sin θ   sin (πa sin θ / λ ) 
I = Imax cos 


λ

  πa sin θ / λ 
2
2
Without the diffraction envelope all of the fringes would have equal intensity. This is what we were
shown in chapter 37, although this pattern doesn’t really exist.
So for this problem, you use the full expression to find the intensity of the bright fringes.
The single slit minima DOMINATE the double slit interference so if a double maxima falls
on a single slit minima, the spot will be dark.
To find the location of the fringes:
Double-slit interference maxima are at angles given by d sin θ
Single-slit diffraction minima, at asin θ = mλ .
= mλ .
(a) The ‘semicylindrical screen’ is irrelevant. The point is, they include that because it would be difficult to SEE
fringes at large angles unless the screen was curved – but it doesn’t matter when you are solving for the
angles.
Double-slit interference maxima are at angles given by d sin θ = mλ .
θ 0=
For m = 0 ,
0°
For m = 1 , ( 2.80 µ m ) sin θ = 1( 0.501 5 µ m ) =
: θ1 sin −1 ( 0.179
=
)
Similarly, for m = 2, 3, 4, 5 and 6,
10.3°
θ2
=
, θ3
21.0° =
, θ4
32.5° =
=
θ5
−1
, and θ 6 sin
63.6°=
=
( 1.07 )
45.8°
nonexistent
Thus, counting the central fringe, there are 5 + 5 + 1 =
11 directions for interference maxima .
(b)
We check for missing orders by looking for single-slit diffraction minima, at asin θ = mλ because
the single slit diffraction DOMINATES the double slit interference:
For m = 1 ,
( 0.700 µ m ) sin θ = 1( 0.501 5 µ m )
and
=
θ1 45.8°
Thus, there is no bright fringe at this angle. There are only nine bright fringes , at
θ = 0°, ± 10.3°, ± 21.0°, ± 32.5°, and ± 63.6° .
c)
To find the ratio of intensities, you have to use this:
 πd sin θ   sin (πa sin θ / λ ) 
I = Imax cos 


λ

  πa sin θ / λ 
2
2
The teacher’s solution is using this equation too but they don’t explain HOW they use it. That is why
reverse engineering the teacher’s solution manual can be a waste of time. Let me explain their
solution. In part (a) we found angles for which d sin θ = mλ . If you put those angles into this equation, the
cosine squared term will always be equal to 1. We solved for the maxima. Where the cosine squared term is
equal to 1. SO that term drops out and we just end up with using:
I = I max
 sin (π a sin θ λ ) 


 π a sin θ λ 
2
Now what they are doing for the rest of the solution is just solving for the argument and making the units work
out because you can’t have use both radian and degrees at the same time in your calculator. Check it out. If I had
just one big equation, see what happens. I’m going to substitute in the givens and simplify, leaving only the
angle:
I
I max
(
)
2
2
 sin π (.7 x10−6 m) sin θ (.5015 x10−6 m) 
 sin ( 4.38507 sin θ ) 
=



 π (.7 x10−6 m) sin θ (.5015 x10−6 m) 
 4.38507 sin θ 


Check it out – if you put in an angle, the top argument in the sine is then in radians and you can’t have
both radians and angles at the same time in your calculator. So the easiest way I THINK to do this
is to make a substitution:
I
I max
 sin ( β ) 
=

 β 
2
where β = π a sin θ λ
Then for each angle, solve for β in degree measure and then switch your calculator to radian and put
in and solve for the final answer.
At θ = 0° ,
lim
sin θ
θ
→ 1 and
β
I
→ 1.00
I max
π a sin θ π ( 0.700 µ m ) sin10.3°
At=
= 0.78406 which is RADIANS.
β=
θ 10.3° , use degrees: =
0.501 5 µ m
λ
Now switch your calculator to radians:
 sin ( 0.78406 ) 
I
= =
0.8107

I max  0.78406 
2
OR, if you don’t want to switch to radians, then you have to convert
keep your calculator in degrees:
β
into degrees for the top and then you can
π asin θ π ( 0.700 µ m ) sin 10.3°
=
= 0.785 rad
= 45.0°
0.501 5 µ m
λ
2
I
 sin 45.0° 
= =
I max  0.785 
0.811
And that is what the teacher solution manual did. It works either way but for this intensity formula you
can’t use both radians and degrees at the same time!!!! So, moral of the story, maybe it is better not to have
the teacher solution manual!!!!
3.
Similarly, at=
θ 21.0° ,
π asin θ
I
= 0.405
= 1.57 rad
= 90.0° and
I max
λ
At=
θ 32.5° ,
π asin θ
I
= 0.090 1
= 2.36 rad
= 135° and
I max
λ
At=
θ 63.6° ,
π asin θ
I
= 0.032 4
= 3.93 rad
= 225° and
I max
λ
The pupil of a cat’s eye narrows to a vertical slit of width 0.500 mm in daylight. What is the
angular resolution for horizontally separated mice? Assume that the average wavelength of the
light is 500 nm.
λ 5.00 × 10−7 m
= 1.00 × 10−3 rad
sin θ= =
−4
5.00 × 10
a
4. The Impressionist painter Georges Seurat
created paintings with an enormous number of
dots of pure pigment, each of which was
approximately 2.00 mm in diameter. The idea
was to have colors such as red and green next to
each other to form a scintillating canvas. Outside
what distance would one be unable to discern
individual dots on the canvas? (Assume that λ =
500 nm and that the pupil diameter is 4.00 mm.)
Sunday Afternoon on the Island of La Grande Jatte,
by Georges Seurat.
By Rayleigh’s criterion, two dots separated center-to-center by 2.00 mm would overlap
when
Thus,
λ
d
= 1.22 .
L
D
2.00
× 10−3 m 4.00 × 10−3 m
dD
=
L =
=
1.22λ
1.22 500 × 10−9 m
θ min=
(
(
)(
)
)
13.1 m .
5. A circular radar antenna on a Coast Guard ship has a diameter of 2.10 m and radiates at a
frequency of 15.0 GHz. Two small boats are located 9.00 km away from the ship. How close
together could the boats be and still be detected as two objects?
1.22
P38.18
λ
D
=
d
L
D = 2.10 m
λ=
c
= 0.020 0 m
f
L = 9 000 m
( 0.020 0 m )( 9 000 m )
=
d 1.22
=
2.10 m
105 m
6. The hydrogen spectrum includes a red line at 656 n and a blue-violet line at 434 nm. What are the
angular separations between these two spectral lines obtained with a diffraction grating that has 4
500 grooves/cm?
=
d
P38.21The grating spacing is
1.00 × 10−2 m
= 2.22 × 10−6 m
4 500
In the 1st-order spectrum, diffraction angles are given by
λ
656 × 10−9 m
sin θ=
= : sin θ1 =
0.295
d
2.22 × 10−6 m
=
θ1 17.17° and for blue
=
sin θ 2
so that for red
so that
434 × 10−9 m
= 0.195
2.22 × 10−6 m
=
θ 2 11.26°
The angular separation is in first-order,
=
∆θ 17.17° − 11.26
=
°
In the second-order spectrum,
2λ
2λ 
∆=
θ sin −1  1  − sin −1  2=

 d 
 d 
13.2°
Again, in the third order,
3λ
3λ 
∆=
θ sin −1  1  − sin −1  2=

 d 
 d 
26.5°
5.91°
Since the red does not appear in the fourth-order spectrum, the answer is complete.
7.
A helium–neon laser (λ = 632.8 nm) is used to calibrate a diffraction grating. If the first-order
maximum occurs at 20.5°, what is the spacing between adjacent grooves in the grating?
P38.22 sin θ = 0.350 :
d
=
λ
=
sin θ
632.8 nm
= 1.81 × 103 nm
0.350
Line spacing = 1.81 µm
8. A diffraction grating has 4 200 rulings/cm. On a screen 2.00 m from the grating, it is found that for
a particular order m, the maxima corresponding to two closely spaced wavelengths of sodium
(589.0 nm and 589.6 nm) are separated by 1.59 mm. Determine the value of m.
1
P38.29 d =
2.38 × 10−6 m =
2 380 nm
=
4 200 cm
d sin θ = mλ
or
mλ 

 d 
θ = sin −1 
and

 mλ  
=
y L=
tan θ L tan sin −1 

 d  




 mλ  
 mλ   
=
∆y L tan sin −1  2   − tan sin −1  1   
 d 
 d  



Thus,
For m = 1 ,
=
∆y












 −1  589   
 589.6  
=
  − tan sin 
   0.554 mm

 2 380  
 2 380   
( 2.00 m ) tan sin −1 
For m = 2 ,
=
∆y
 −1  2 ( 589)   
 2 ( 589.6)  
=
  − tan sin 
   1.54 mm

 2 380  
 2 380   
( 2.00 m ) tan sin −1 
For m = 3 ,
=
∆y
 −1  3 ( 589)   
 3 ( 589.6)  
=
  − tan sin 
   5.04 mm

 2 380  
 2 380   
( 2.00 m ) tan sin −1 
Thus, the observed order must be m = 2 .
9. Unpolarized light passes through two ideal Polaroid sheets.The axis of the first is vertical and that
of the second is at 30.0° to the vertical. What fraction of the incident light is transmitted?
P38.34 The average value of the cosine-squared function is one-half, so the first polarizer transmits
3
transmits cos2 30.0° = .
4
1
the light. The second
2
1 3
3
I f = × Ii = Ii
2 4
8
10. The critical angle for total internal reflection for sapphire surrounded by air is 34.4°. Calculate the
polarizing angle for sapphire.
sin θ c =
P38.37
Also, tan θ p = n .
1
n
or
=
n
1
1
=
= 1.77 .
sin θ c sin 34.4°
Thus, =
θ p tan −1=
=
( n) tan −1 ( 1.77
)
60.5° .
11. The Very Large Array (VLA) is a set of 27 radio telescope
dishes in Caton and Socorro Counties, New Mexico
(Fig. P38.54). The antennas can be moved apart on
railroad tracks, and their combined signals give the
resolving power of a synthetic aperture 36.0 km in
diameter. (a) If the detectors are tuned to a frequency of
1.40 GHz, what is the angular resolution of the VLA?
(b) Clouds of hydrogen radiate at this frequency. What
must be the separation distance of two clouds at the
center of the galaxy, 26 000 lightyears away, if they are
to be resolved? (c) What If? As the telescope looks up,
a circling hawk looks down. Find the angular
resolution of the hawk’s eye. Assume that that the
hawk is most sensitive to green light having a wavelength of 500 nm and that it has a pupil of diameter 12.0
mm. (d) A mouse is on the ground 30.0 m below. By what distance must the mouse’s whiskers be separated
if the hawk can resolve them?
P38.50
(a)
v
f
λ=
= :
λ
3.00 × 108 m s
=
0.214 m
1.40 × 109 s−1
λ
 0.214 m 
:
θ min 1.22
=
=
θ min = 1.22


D
3.60 × 104 m 
 180 × 60 × 60 s
=
=
θ min 7.26
µrad 


π
(
7.26 µrad
1.50 arc seconds
)
d=
7.26 × 10−6 rad ( 26 000 ly ) =
0.189 ly
θ min L =
d
:
L
(b)
θ min =
(c)
 500 × 10−9 m 
λ
=
=
θ min = 1.22 θ min 1.22


D
 12.0 × 10−3 m 
(d)
θ min L =
d=
50.8 × 10−6 rad ( 30.0 m ) =
1.52 × 10−3 m =
1.52 mm
(
50.8 µrad ( 10.5 seconds of arc)
)
12. An American standard television picture is composed of about 485 horizontal lines of varying light intensity.
Assume that your ability to resolve the lines is limited only by the Rayleigh criterion and that the pupils of
your eyes are 5.00 mm in diameter. Calculate the ratio of minimum viewing distance to the vertical
dimension of the picture such that you will not be able to resolve the lines. Assume that the average
wavelength of the light coming from the screen is 550 nm.
(
(
)
)
550 × 10−9 m
λ
P38.57 The limiting resolution between lines θ min
= 1.22= 1.22
= 1.34 × 10−4 rad .
D
5.00 × 10−3 m
θ min
Assuming a picture screen with vertical dimension  , the minimum viewing distance for no visible lines is found from
 485
. The desired ratio is then
=
L
L
1
1
= =
=
 485θ min 485 1.34 × 10−4 rad
(
)
15.4 .
13. If the interplanar spacing of NaCl is 2.814 × 10–10 m, what is the predicted angle at which X-rays of
wavelength 0.140 nm will be diffracted in a first-order maximum?
Answer: 14.4°
14. Suppose a 5-meter diameter telescope were constructed on the dark side of the moon. The viewing there
(except for brief periods of sunlight) would be excellent. As an example, what would be the separation
between two objects that could just be resolved on the planet Mars in 500 nm light? [The distance to Mars at
closest approach is 50 million miles].
Answer: 9.8 km
15. Suppose a 5-meter diameter telescope were constructed on the dark side of the moon. The viewing there (except for brief
periods of sunlight) would be excellent. As an example, what would be the separation between two objects that could just be
resolved on the planet Mars in 500 nm light? [The distance to Mars at closest approach is 50 million miles].
16. An unpolarized beam of light is incident on a stack ofideal polarizing filters. The axis of the first filter is
perpendicular to the axis of the last filter in the stack. Find the fraction by which the transmitted beam’s intensity is
reduced in the following three cases. (a) Three filters are in the stack, each with its transmission axis at 45.0° relative
to the preceding filter. (b) Four filters are in the stack, each with its transmission axis at 30.0° relative to the
preceding filter. (c) Seven filters are in the stack, each with its axis at 15.0° relative to the preceding filter.(d)
Comment on comparing the answers to parts (a), (b), and (c).
17. Monochromatic light is beamed into a Michelson interferometer. The movable mirror is displaced 0.382 mm,
causing the interferometer pattern to reproduce itself 1 700 times. Determine the wavelength of the light. What color
is it?