The previous section of Lesson 3 discussed Thomas Young's
effort to derive an equation relating the wavelength of a light source to
reliably measured distances associated with a two-point source light
interference pattern. The equation, known as Young's equation is:
λ =
y • d / (m • L)
In 1801, Young devised and performed an experiment to measure
the wavelength of light. As discussed in the previous part of this lesson, it was important that the two sources of light that
form the pattern be coherent. The difficulty confronting Young was that the
usual light sources of the day (candles, lanterns, etc.) could not serve as
coherent light sources. Young's method involved using sunlight that entered the
room through a pinhole in a window shutter. A mirror was used to direct the
pinhole beam horizontally across the room. To obtain two sources of light,
Young used a small paper card to break the single pinhole beam into two beams,
with part of the beam passing by the left side of the card and part of the beam
passing by the right side of the card. Since these two beams emerged from the
same source - the sun - they could be considered coming from two coherent sources.
Light waves from these two sources (the left side and the right side of the
card) would interfere. The interference pattern was then projected onto a
screen where measurements could be made to determine the wavelength of light.
Today's classroom version of the same experiment is typically
performed using a laser beam as the source. Rather than using a note card to
split the single beam into two coherent beams, a carbon-coated glass slide with
two closely spaced etched slits is used. The slide with its slits is most
commonly purchased from a manufacturer who provides a measured value for the
slit separation distance - the d value in Young's equation. Light from the laser
beam diffracts through the slits and emerges as two separate coherent waves.
The interference pattern is then projected onto a screen where reliable
measurements can be made of L and y for a given bright spot with order value m. Knowing these four values allows a
student to determine the value of the wavelength of the original light source.
To illustrate some typical results from this experiment and
the subsequent analysis, consider the sample data provided below for d, y, L and m.
Data Table |
|
Slit Separation (d) |
0.250 mm |
Distance from Slits to Screen (L) |
9.78 m |
Distance from AN0 to AN4(y) |
10.2 cm |
Order value (m) |
4 |
(Note: AN0 = central antinode
and AN4 = fourth antinode)
The determination of the wavelength demands that the above
values for d, y, L and m be substituted into Young's equation.
λ =
y • d / (m • L)
Careful inspection of the units of measurement is always
advisable. The sample data here reveal that each measured quantity is recorded
with a different unit. Before substituting these measured values into the above
equation, it is important to give some thought to the treatment of units. One
means of resolving the issue of nonuniform units is to simply pick a unit of length
and to convert all quantities to that unit. If doing so, one might want to pick
a unit that one of the data values already has so that there is one less
conversion. A wise choice is to choose the meter as the unit to which all other
measured values are converted. Since there are 1000 millimeters in 1 meter, the 0.250 mm is equivalent to 0.000250 meter. And
since there are 100 centimeters in 1 meter, the 10.2 cm is equivalent to 0.102 m. Thus, the new
values of d, y and L are:
d |
0.000250 m |
L |
9.78 m |
y |
0.102 m |
m |
4 |
While the conversion of all the data to the same unit is not
the only means of treating such measured values, it might be the most advisable
- particularly for those students who are less at ease with such conversions.
Now that the issue regarding the units of measurement has
been resolved, substitution of the measured values into Young's equation can be
performed.
λ =
(0.102 m) • (0.000250 m) / [ (4) • (9.78 m) ]
λ = 6.52 x 10-7 m
As is evident here, the wavelength of visible light is rather
small. For this reason wavelength is often expressed using the unit nanometer, where 1
meter is equivalent to 109 nanometers. Multiplying by 109will convert the wavelength from meters
to nanometers (abbreviated nm).
λ =
652 nm
1. The diagram below depicts the results of Young's
Experiment. The appropriate measurements are listed on the diagram. Use these
measurements to determine the wavelength of light in nanometers. (GIVEN: 1
meter = 109 nanometers)
Answer: 657 nm
First, identify known values in terms of their corresponding
variable symbol:
L = 10.2 m = 1020 cm |
y = 22.5 cm |
m = 10 |
d = 0.298 mm = 0.0298 cm |
(Note: m was chosen as 10 since the y distance corresponds to
the distance from the 5th bright band on one side of the central band and the
5th bright band on the other side of the central band.)
Then convert all known values to an identical unit. In this
case, cm has been chosen as the unit to use. The converted values are listed in
the table above.
Substitute all values into Young's equation and perform
calculation of the wavelength. The unit of wavelength is cm.
λ = y • d / ( m • L)
λ = ( 22.5 cm ) • ( 0.0298 cm ) / [ ( 10 ) • ( 1020 cm ) ]
λ = 6.57 x 10-5 cm
Finally convert to nanometers using
a conversion factor. If there are 109 nm in 1 meter, then there
must be 107 nm in the smaller centimeter.
λ = ( 6.57 x 10-5 cm
) • ( 107 nm / 1 cm ) = 657 nm
2. A student uses a laser and a double-slit apparatus to
project a two-point source light interference pattern onto a whiteboard located
5.87 meters away. The distance measured between the central bright band and the
fourth bright band is 8.21 cm. The slits are separated by a distance of 0.150
mm. What would be the measured wavelength of light?
Answer: 524 nm
First, identify known values in terms of their corresponding
variable symbol:
L = 5.87 m = 587 cm |
y = 8.21 cm |
m = 4 |
d = 0.150 mm = 0.0150 cm |
Then convert all known values to an identical unit. In this
case, cm has been chosen as the unit to use. The converted values are listed in
the table above.
Substitute all values into Young's equation and perform
calculation of the wavelength. The unit of wavelength is cm.
λ = y • d / ( m • L)
λ = ( 8.21 cm ) • ( 0.0150 cm ) / [ ( 4 ) • ( 587 cm ) ]
λ = 5.24 x 10-5 cm
Finally convert to nanometers using
a conversion factor. If there are 109 nm in 1 meter, then there
must be 107 nm in the smaller centimeter.
λ = ( 5.24 x 10-5 cm
) • ( 107 nm / 1 cm ) = 524 nm
3. The analysis of any two-point source interference pattern
and a successful determination of wavelength demands an ability to sort through
the measured information and equating the values with the symbols in Young's
equation. Apply your understanding by interpreting the following statements and
identifying the values of y, d, m and L. Finally, perform some conversions of
the given information such that all information share the same unit.
a. Two slits separated by 0.250 mm produces an
interference pattern in which the fifth dark band is located 12.8 cm from the
central antinode when the screen is placed a distance of 8.2 meters away.
y = |
d = |
m = |
L = |
This question simply asks to equate the stated information with
the variables of Young's equation and to perform conversions such that all
information is in the same unit.
y = 12.8 cm |
d = 0.250
mm |
m = 4.5 |
L = 8.2
meters |
Converted values: |
|||
y = 12.8 cm |
d = 0.0250
cm |
m = 4.5 |
L = 820 cm |
(Note that m = 4.5 represents the fifth nodal position or dark
band from the central bright band. Also note that the given values have been
converted to cm.)
b. An interference pattern is produced when light is incident
upon two slits that are 50.0 micrometers apart. The perpendicular distance from
the midpoint between the slits to the screen is 7.65 m. The distance between
the two third-order antinodes on opposite sides of the pattern is 32.9 cm.
y = |
d = |
m = |
L = |
This question simply asks to equate the stated information with
the variables of Young's equation and to perform conversions such that all
information is in the same unit.
y = 32.9 cm |
d = 50.0 µm |
m = 6 |
L = 7.65 m |
Converted values: |
|||
y = 32.9 cm |
d =
0.00500 cm |
m = 6 |
L = 765 cm |
(Note that m = 6 corresponds to six spacings.
There are three spacings between the
central antinode and the third antinode. The stated distance is twice as far so
the m value must be doubled. Also note that the given values have been
converted to cm. There are 106 µm in one meter; so there are 104 µm
in one centimeter.)
c. The fourth nodal line on an interference pattern is 8.4 cm
from the first antinodal line when the screen is placed 235 cm from the slits. The slits
are separated by 0.25 mm.
y = |
d = |
m = |
L = |
This
question simply asks to equate the stated information with the variables of
Young's equation and to perform conversions such that all information is in the
same unit.
y = 8.4 cm |
d = 0.25 mm |
m = 2.5 |
L = 235 cm |
Converted values: |
|||
y = 8.4 cm |
d = 0.025 cm |
m = 2.5 |
L = 235 cm |
( Note that the fourth nodal line is assigned the order
value of 3.5. Also note that the given values have been converted to cm.)
d. Two sources separated by 0.500 mm produce an interference
pattern 525 cm away. The fifth and the second antinodal line on the same side of the pattern are
separated by 98 mm.
y = |
d = |
m = |
L = |
This question simply asks to equate the stated information with
the variables of Young's equation and to perform conversions such that all
information is in the same unit.
y = 98 mm |
d = 0.500
mm |
m = 3 |
L = 525 cm |
Converted values: |
|||
y = 9.8 cm |
d = 0.0500
cm |
m = 3 |
L = 525 cm |
( Note that there are
three spacings between the second and the
fifth bright bands. Since all spacings are
the same distance apart, the distance between the second and the fifth bright
bands would be the same as the distance between the central and the third
bright bands. Thus, m = 3. Also note that the given values have been converted
to cm.)
e. Two slits that are 0.200 mm apart produce an interference
pattern on a screen such that the central maximum and the 10th bright band are
distanced by an amount equal to one-tenth the distance from the slits to the
screen.
y = |
d = |
m = |
L = |
This
question simply asks to equate the stated information with the variables of
Young's equation and to perform conversions such that all information is in the
same unit.
y
= 0.1 • L |
d
= 0.200 mm |
m
= 10 |
L
- not stated |
Converted values: |
|||
y
= 0.1 • L |
d
= 0.200 mm |
m
= 10 |
L
- not stated |
( Note that there are 10 spacings between the central anti-node and the
tenth bright band or tenth anti-node. And observe that they do not state the
actual values of L and y; the value of y is expressed in terms of L. )
f. The fifth antinodal line and the second nodal line on the
opposite side of an interference pattern are separated by a distance of 32.1 cm
when the slits are 6.5 m from the screen. The slits are separated by 25.0 micrometers.
y = |
d = |
m = |
L = |
This question simply asks to equate the stated information with
the variables of Young's equation and to perform conversions such that all
information is in the same unit.
y = 32.1 cm |
d = 25.0 µm |
m = 6.5 |
L = 6.5 m |
Converted values: |
|||
y = 32.1 cm |
d =
0.00250 cm |
m = 6.5 |
L = 650 cm |
( Note that there are
five spacings between the central anti-node
and the fifth anti-node. And there are 1.5 spacings from
the central anti-node in the opposite direction out to the second nodal line.
Thus, m = 6.5. Also note that the given values have been converted to cm. There
are 106 µm in one meter; so there are 104 µm in
one centimeter.)
g. If two slits 0.100 mm apart are separated from a screen by
a distance of 300 mm, then the first-order minimum will be 1 cm from the
central maximum.
y = |
d = |
m = |
L = |
This question simply asks to equate the stated information with
the variables of Young's equation and to perform conversions such that all
information is in the same unit.
y = 1 cm |
d = 0.100
mm |
m = 0.5 |
L = 300 mm |
Converted values: |
|||
y = 1 cm |
d = 0.0100
cm |
m = 0.5 |
L = 30.0 cm |
( Note that a the
first-order minimum is a point of minimum brightness or a nodal position. The
first-order minimum is the first nodal position and is thus the m = 0.5 node.
Also note that the given values have been converted to cm. )
h. Consecutive bright bands on an interference pattern are
3.5 cm apart when the slide containing the slits is 10.0 m from the screen. The
slit separation distance is 0.050 mm.
y = |
d = |
m = |
L = |
This question simply asks to equate the stated information with
the variables of Young's equation and to perform conversions such that all
information is in the same unit.
y = 3.5 cm |
d = 0.050
mm |
m = 1 |
L = 10.0 m |
Converted values: |
|||
y = 3.5 cm |
d = 0.0050
cm |
m = 1 |
L = 1000 cm |
( Note that the
spacing between adjacent bands is given. This distance is equivalent with the
distance from the central bright band to the first antinode. Thus, m = 1. Also
note that the given values have been converted to cm. )