University of Rochester
Department of Physics and Astronomy
Physics123, Spring 2012
Recitation 7 - Solutions
Conceptual Questions:
• We can hear sounds around corners but we cannot see around corners; yet both sound
and light are waves. Explain the difference.
The bending of waves around corners or obstacles is called diffraction (see 34-1).
Diffraction is most prominent when the size of the obstacle is on the order of the size of
the wavelength. Sound waves that we can hear have much longer wavelengths than do
light waves. As a result, the diffraction of sound waves around a corner is noticeable and
we can hear the sound in the “shadow region,” but the diffraction of light waves around a
corner is not noticeable. Since Huygens’ principle applies to all waves, you may see this
effect in figure 34-2a.
• Explain why two flashlights held close together do not produce an interference pattern
on a distant screen.
The two flashlights are two incoherent sources. They consist of many different
wavelengths (that is why they are white) with random time differences between the light
waves. The light from the two flashlights does not maintain a constant phase relationship
over time.
• If double-slit experiment were submerged in water, how would the fringe pattern be
changed?
The fringes would be closer together because the wavelength of the light underwater is
less than the wavelength in air.
• Some coated lenses appear greenish yellow when seen by reflected light. What
wavelengths do you suppose the coating is designed to eliminate completely?
These lenses probably are designed to eliminate wavelengths at both the red and the blue
ends of the spectrum. The thickness of the coating is designed to cause destructive
interference for reflected red and blue light. The reflected light then appears yellowgreen.
Problem R7.1
In a double-slit experiment the aperture-to-screen distance (L) is 2.0 m and the
wavelength of the incident light is 600 nm.
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(a) If it is desired to have a first and second order fringe spacing of 1 mm, what is d, the
required slit separation?
A glass plate is placed in front of the lower slit so that the two waves enter the slits π
out of phase (see figure).
(b) Describe in detail the interference pattern on the screen.
(c) If the glass plate has refractive index n = 1.5 and the wavelength of the incident light
is 600 nm, calculate the thickness of the plate that produces the π out of phase shift.
Solution R7.1
(a) Following Giancoli, Example 34-2
θ ≅ sin θ =
mλ
;
d
θ1 ≅
λ
x2 − x1 ≅ L(θ 2 − θ1 ) = L
d
λ
d
λ
θ2 ≅ 2 ;
d
⇒
→
x1 ≅ Lθ1
x2 ≅ Lθ 2
Lλ
(2 m)(600 ×10 −9 m)
d≅
=
= 1.2 ×10 −3 m
−3
x2 − x1
10 m
(b) The π phase shift produced by the glass in figure 1 is equivalent to a path length of
1
λ.
2
For constructive interference on the screen, the total path difference is a multiple of the
wavelength:
1
2
λ + d sin θ max = mλ , m = 0, 1, 2,  → d sin θ max = ( m − 12 ) λ , m =1, 2, 
We could express the result as d sin θ max = ( m + 12 ) λ , m = 0, 1, 2, .
For destructive interference on the screen, the total path difference is
1
2
λ + d sin θ min = ( m + 12 ) λ , m = 0, 1, 2, → d sin θ min = mλ , m = 0, 1, 2,
Thus the pattern is just the reverse of the usual double-slit pattern. There will be a dark
central line. Every place there was a bright fringe will now have a dark line, and vice
2
versa.
(c) The π phase difference is given by
k2 l
l=
2⇡
k1 l = ⇡;
nl
2(n 1) = 600 nm
2⇡
l=
2⇡
l(n
1) = ⇡
where l is the glass thickness.
Problem R7.2
Let us explore why only “thin” layers exhibit thin-film interference when illuminated by
white light. Assume a layer of water, sitting atop a flat glass surface (nglass = 1.50), is
illuminated from the air above by white light (all wavelengths from 400 nm to 700 nm).
Further, assume that the water layer’s thickness t is 200 µm, that is, much greater than
visible wavelengths. Take the index of refraction for water to be constant (n = 1.33) for
all visible wavelengths.
(a) Show that a visible color will be reflected from the water layer if its wavelength is
λ = 2nt m , where m is an integer.
(b) Show that the two extremes in wavelength (400 nm and 700 nm) of the incident light
are both reflected from the water layer and determine the m-value associated with each.
(c) How many other visible wavelengths, besides 400 nm and 700 nm, are reflected
from the “thick’ layer of water?
(d) How does this explain why such a tick layer does not reflect colorfully, but is white
or grey?
Problem R7.2 – Solutions
φ1 = π
(a)
Assume the indices of refraction for air,
water, and glass are 1.00, 1.33, and 1.50,
φ2 = ( 2t λfilm ) 2π + π
respectively. When illuminated from above, a ray air
reflected from the air-water interface undergoes a
water
phase shift of φ1 = π , and a ray reflected at the waterglass
glass interface also undergoes a phase shift of π .
Thus, the two rays are unshifted in phase relative to each other due to reflection. For
constructive interference, the path difference 2t must equal an integer number of
wavelengths in water.
λ
2n t
2t = mλwater = m
, m = 0, 1, 2, L → λ = water
nwater
m
3
(b)
The above relation can be solved for the m-value associated with the reflected
color. If this m-value is an integer the wavelength undergoes constructive interference
upon reflection.
2nwater t
2n t
→ m = water
m
λ
For a thickness t = 200µ m = 2 × 105 nm the m-values for the two wavelengths are
λ=
calculated.
m700 nm =
m400 nm =
2nwater t
λ
2nwater t
=
=
2 (1.33) ( 2 × 105 nm )
700nm
2 (1.33) ( 2 × 105 nm )
= 760
= 1330
λ
400nm
Since both wavelengths yield integers for m, they are both reflected.
(c)
All m-values between m = 760 and m = 1330 will produce reflected visible colors.
There are 1330 – (760 – 1) = 571 such values.
(d)
This mix of a large number of wavelengths from throughout the visible spectrum
will not select any specific color but will give the thick layer a white or grey appearance.
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