PHYSICS 6C
Ch.26 Worksheet Solutions
1) Two students in a dorm room listen to a pure tone produced by two
speakers that are in phase. Both students hear a maximum sound. What is
the lowest possible frequency of the sound? (speed of sound=343 m/s)
Student A is equal distance from each
speaker, so as long as they are in
phase, he will hear a maximum sound.
3.35m
For student B, he is 1.5m from the
close speaker, and 3.35m from the far
speaker (Pythagorean thm). The
difference is 1.85m.
This is the longest wavelength that will
produce constructive interference at his
location.
Using 343 m/s for the speed of sound,
we get a frequency of 185Hz.
2) Light from a He-Ne laser (λ=632.8nm) strikes a pair of slits at normal
incidence, forming a double-slit interference pattern on a screen located
1.40m from the slits, as shown. What is the slit separation?
The 23mm distance covers 4 fringes, so
we can use m=4 in the formula for fringe
separation:
ym = R
(4)(632.8 ⋅ 10 −9 m)
mλ
⇒ d = (1.4m)
d
23 ⋅ 10 − 3 m
d = 1.54 ⋅ 10 − 4 m
3) When green light (λ=505nm) passes through a pair of slits, the
interference pattern (a) is observed. When light of a different color passes
through the same pair of slits, pattern (b) is observed instead.
a) Is the wavelength of the second color greater or less than 505nm?
b) Find the wavelength of the second color.
We can use the formula for fringe
locations for each case:
(4.5)(505nm)
for (a): y = R
d
(5)(λ)
for (b): y = R
d
Set these equal and solve for λ.
4.5
(505nm) = 454.5nm
λ =
5
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PHYSICS 6C
Ch.26 Worksheet Solutions
4) An air wedge is formed by placing a human hair between two glass plates
on one end, and allowing them to touch on the other end. When this wedge
is illuminated with red light (λ=771nm), it is observed to have 179 dark
fringes. How thick is the hair? This ray is shifted
This ray is not shifted
because it reflects from a
lower index (glass to air)
because it reflects
from a higher index
(air to glass)
The 2 rays that are interfering have
a relative phase shift, so they are
already out of phase. Thus for dark
fringes we use the formula 2t=mλ.
Set m=179 and solve for t:
t=
(179)(771 ⋅ 10 −9 m)
= 6.9 ⋅ 10 −5 m
2
5) The diffraction pattern shown in the figure is produced by passing He-Ne
laser light (λ=632.8nm) through a single slit and viewing the pattern on a
screen 1.5m behind the slit.
a) What is the width of the slit?
b) If monochromatic yellow light of wavelength 591nm is used with this slit
instead, will the distance in the figure be greater or less than 15.2cm?
The distance from the center to the 2nd dark
fringe is 7.6cm (half of the distance shown).
mλ
with m=2.
Use the formula y m = R
a
part (b) Longer wavelength = wider fringes.
a = (1.5m)
(2)(591 ⋅ 10 −9 m)
= 2.4 ⋅ 10 − 5 m
0.076m
6) White light strikes a diffraction grating with 7400 lines/cm at normal
incidence. How many complete visible spectra will be formed on either side
of the central maximum? (visible light has λ between 400nm and 700nm)
1
cm . Lnger wavelengths
7400
correspond to larger angles, so we use λ = 700nm , complete spectra will appear on the
screen only if the angle given by the formula is smaller than 90°.
We can use the formula d sin(θ) = mλ with slit spacing d =
sin(θ) =
mλ m(700 ⋅ 10 −9 m)
=
= m(0.518)
1
d
7400 cm
If m=2 our formula doesn’t work, so we get 1 full spectrum on either side of the center.
7) The asteroid Ida is orbited by its own small “moon” called Dactyl. If the
separation between these two asteroids is 2.5km, what is the maximum
distance at which the Hubble Space Telescope (aperture diameter 2.4m) can
still resolve them with 550nm light?
Use Rayleigh’s criterion to find the θmin.
λ
550 ⋅ 10 −9 m
= 1.22
= 2.8 ⋅ 10 − 7 rad
2.4m
D
From the triangle in the picture we get
y
2500m
tan(θ) =
⇒L =
= 8.9 ⋅ 10 9 m
L
tan(2.8 ⋅ 10 − 7 rad)
θ min = 1.22
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