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thin film interference

By blesson123 Feb 21, 2014 1509 Words
Experiment 112-5
Thin Film Interference
Introduction
When two waves of equal intensity combine, the intensity of the resulting wave can be anywhere between zero and four times the intensity of the individual waves, depending upon the phase difference between them. When they are in phase (the maxima coincide with maxima and the minima coincide with minima), maximum intensity results. When they are exactly out of phase (the maxima of one wave coincide with the minima of the other wave), the result is zero intensity.

In thin film interference, a beam of light is separated into two parts that travel on different paths and then recombine. If the difference in path lengths is exactly an integral number of wavelengths, the waves recombine in phase, and constructive interference occurs. If the path lengths differ by exactly one half-wavelength, the waves are out of phase when they recombine and destructive interference occurs.

THEORY
Thin film interference can occur when two glass plates are placed in contact at one end and are separated by some small distance t at the other. Consider a monochromatic light wave of wavelength λ striking the glass plates at right angles to their surfaces. Part of the light wave will be reflected back at the boundary between the first glass plate and the space. Another part of the light wave will travel across the space and will be reflected back from the front surface of the second glass plate. These two reflected waves then combine.

The difference in optical path length for the two light waves is (2t − λ/2). The λ/2 term arises because there is a phase change of one half a wavelength when the light is reflected off the second glass plate. This effectively alters the physical path length by λ/2. (This assumes that the space between the two glass plates is filled with air, an optically less dense material than glass.) For constructive interference this difference in path length must equal m wavelengths, where m is an integer. Thus,

mλ = 2t −

λ
2

(5.1)

where m = 0,1,2, etc. Solving equation (5.1) for t gives
t=

mλ λ
+
2 4

(5.2)

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Thin Film Interference

Experiment 112-5

For destructive interference the difference in path length must equal an integral number of wavelengths less a half wavelength. Thus
1
λ

 m −  λ = 2t − .
2
2


(5.3)

Solving equation (5.3) for t gives
t=


.
2

(5.4)
The first example of thin film
interference to be studied in this experiment
is that of an air wedge formed between two
plates of glass when an object of thickness
T is placed between them, as illustrated in
Figure 5.1.
When monochromatic light is
allowed to shine onto the interference
plates, a series of dark bands or fringes is
observed as shown in Figure 5.2.
A dark band is formed at the point
where the two glass plates are in contact
because there is an effective λ/2 difference
in path length between the light waves
reflected from the top and bottom plates due
to the λ/2 phase change upon reflection at
the bottom plate. The next (m = 1) dark
fringe occurs when the interference plates
are separated by a distance of λ/2, the m = 2
fringe occurs at a plate separation of 2(λ/2),
and so on. The mth fringe occurs at a plate
separation of m(λ/2). Consequently, if there
are m dark bands between the object and the
point where the glass plates are in contact,

Figure 5.1 Interference plates

Figure 5.2 Interference Fringes

then the thickness of the object is given by:
T=


2

(5.5)

If the wavelength of the incident light is known, the thickness of an object placed between two interference plates can be found simply by counting the number of dark interference fringes that are formed.

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Thin Film Interference

Experiment 112-5

A second example of thin film
interference is Newton's rings. Newton's rings
occur when monochromatic light is shone on a
lens that has been placed in contact with a glass
plate as shown in Figure 5.3.
Newton’s own drawing of the pattern of
rings from his book Opticks is shown in the
bottom part of Figure 5.3. Note that at the
center, where the lens is in contact with the glass
plate, a dark circle is formed. This occurs for the
same reason that a dark fringe occurs at the point
where the two interference plates are in contact
in the previous example. Also, using reasoning
similar to that used in the previous example, it
can be seen that the separation s between the
bottom of the lens and the glass plate at the
position of mth dark ring is given by
s=


2

(5.6)

This can be used to determine the radius of
curvature, R of the lens in a manner similar to
that which was used with the spherometer in
Experiment 3.

Figure 5.3 Newton’s rings

PURPOSE
The object of the first part of the experiment is to determine the thickness of a human hair by observing how many interference fringes are formed when it is placed between two interference plates.

The object of the second part of the experiment is to use Newton’s rings to determine the radius of curvature of both sides of a lens. The focal length of the lens will also be determined by using the lens maker’s formula.

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Thin Film Interference

Experiment 112-5

PROCEDURE
Carefully clean the interference
plates and the lens with lens tissue
before trying to form interference
fringes and Newton’s rings.
1. Thickness of a Hair
Set up the apparatus as shown in
Figure 5.4. A sodium lamp is used
to illuminate the interference plates.
Put a piece of black felt cloth under
the plates on the microscope stage.
A sloping glass plate placed
between the interference plates and
the microscope optics will direct the
light directly onto the plates from
above. Place a human hair near one
end between the upper and lower
interference plates and observe the
fringes through the microscope.
Figure 5.4 General set-up

Before attempting to find the fringes and rings in the measuring microscope, remove the microscope tube and find the fringes and rings by looking down the cylindrical microscope holder with your naked eye. Adjust the alignment of the interference plates, lens and reflection glass plate until the fringes or rings are clearly visible to your eye. If you cannot clearly see them with your naked eye you will probably never find them in the microscope. Count the dark fringes between the hair and the point of contact between the two plates. If the number of fringes is too large to count conveniently, it may be determined as follows. Use the measuring microscope to determine the distance between ten (or twenty) fringes, then calculate the number of fringes per unit length. Next, use the measuring microscope to measure the distance between the hair and the point at which the two glass plates come into contact. The product of the distance and the fringe density gives the total number of fringes. Look up the wavelength of sodium light in the Handbook of Chemistry and Physics (or some other suitable reference). The yellow light from the lamp is due to what are called the sodium D lines, a close doublet of spectral lines. You should use the intensity weighted average [i.e., λ = ( I1λ1 + I 2λ2 ) ( I1 + I 2 ) ] of the wavelengths in your analysis. Then use Equation (5.5) to calculate the thickness of the hair.

Use a micrometer caliper to measure the thickness of the hair as a check on your results. Compare the two values.

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Thin Film Interference

Experiment 112-5

2. Newton’s Rings
Set up the apparatus as shown in Figure 5.4, but substitute the long focal length for the hair and the upper interference plate. Observe the rings through the measuring microscope.

The center of the ring pattern should be dark. If it isn’t, there is a small air space between the lens and the glass plate. Clean both and try again. If, after many tries, you just cannot get the center ring dark, count bright rings instead of dark rings.

Use the measuring microscope to measure the diameter of the tenth dark ring. (Remember, the center ring counts as the zeroth ring.) Calculate the distance s between the lens and the glass plate via Equation (5.6), then apply Pythagoras’ Theorem to calculate the radius of curvature R of the lens. (See Figure 5.3.)

Reverse the lens and use the same method to find the radius of curvature of the second side. Then use the lens maker’s formula [Equation (3.3)] to calculate the focal length f of the lens. The index of refraction of the glass given by the manufacturer of the lens is 1.523. Check the value of the focal length of the lens by setting up the lens to form an image with apparatus set up on the side bench in the laboratory via the method to be used in Experiment 3. You only need to do one set of image and object distance measurements. Use Equation (3.1) to calculate the focal length.

Would it have been possible to do the Newton’s rings part of the experiment with the flat glass plate sitting on top of the lens rather than the way it was done? Explain.

25

Thin Film Interference

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Experiment 112-5

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