A
disabled tanker leaks kerosene
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Solution: Click For PDF Version (a)
It is given
that the thickness of the kerosene slick
where x is the thickness of the kerosene slick, 460 nm. We know that
where
Waves (1) and
(2) will interfere constructively, if their phase difference is
Therefore, the waves with the following wavelengths will appear brightest to an observer who is sighting the slick from vertically above it:
We thus find that the waves of the following wavelengths will appear with greatest intensity when viewed from above the slick:
Out of these we note that the wavelength in the visible region is 552 nm.
(b)
For answering
the second part of the problem we will consider interference of
waves (3) and (4), see the figure. Wave (3) is transmitted from air
to kerosene and then to water. But wave (4) undergoes two
reflections; one from kerosene-water interface and another from
kerosene-air interface. As the refractive index
of sea water is greater than that of kerosene there is an additional
phase change of
The wavelengths for which the transmitted waves in water will appear brightest will be determined by the following condition:
Out of these the wavelength in the visible part of the spectrum is 442 nm.
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into the Persian gulf,
creating a large slick on top of the water
.
(a) If we are looking straight down from an
airplane onto a region of the slick where the thickness is
460 nm, we have to find the wavelength(s)
of the visible light for which the reflection
is the greatest. (b) If
we are scuba-diving directly under the same region of the slick,
we have to find the wavelength(s)
of the transmitted visible light for which the
intensity is strongest.
.
From the figure we note that both waves (1) and (2) undergo one
reflection each from media having greater index of refraction than
the refractive index of the incident media, therefore each acquires
a phase change of
is the wavelength in air.
