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A thin film of \(\mathrm{MgF}_{2}(n=1.38)\) coats a piece of glass. Constructive interference is observed for the reflection of light with wavelengths of \(500 \mathrm{nm}\) and \(625 \mathrm{nm} .\) What is the thinnest film for which this can occur?

Short Answer

Expert verified
The thinnest film for which constructive interference can occur for both light waves is approximately 227 nm in thickness.

Step by step solution

01

Apply the Constructive Interference Condition

To get the minimum thickness that allows constructive interference for the given wavelengths, the next equation will be used with \(m = 1\) (since the minimum occurs when m is 1). Plugging \(n = 1.38\), \(r = 0\) (since it is normal incidence), \(m = 1\), and the provided wavelengths into the formula \(2 \cdot n \cdot d \cdot \cos(r) = m \cdot \lambda\), we obtain: \(2 \cdot 1.38 \cdot d = 1 \cdot \lambda\). Simplifying the expression gives: \(d = \lambda / (2 \cdot 1.38)\). Now, we apply this formula with both given wavelengths.
02

Calculate the Thickness for Each Wavelength

Substitute \(\lambda = 500 nm\) into the formula we derived from Step 1: \(d = \frac {500} {2 \cdot 1.38} = 181 nm\). Similarly, substitute \(\lambda = 625 nm\): \(d = \frac {625} {2 \cdot 1.38} = 227 nm\). So, for the two different wavelengths, we've found two different thicknesses that correspond to the minimum thickness for constructive interference to occur.
03

Determine the Minimum Thickness for Both Wavelengths

Just because the film thickness can create constructive interference for one wavelength doesn't mean it will work for another. The student has to find a common thickness that will create constructive interference for both wavelengths (500 nm and 625 nm). The common thickness occurs when the thickness calculated for the first wavelength, multiplied by an integer, equals the thickness calculated for the second wavelength multiplied by another integer. Upon rationalizing, we'll find out that the smallest thickness for which this happens is approximately 227 nm (calculated previously), which is the minimum thickness of the film that causes constructive interference for both light rays with wavelengths of 500 nm and 625 nm.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constructive Interference
Constructive interference occurs when two or more light waves meet and combine to make a wave of greater amplitude. This happens when their valleys and peaks align. In terms of light and thin films, this results in bright colors.
Constructive interference in thin films happens when the light waves reflecting off the top and bottom surfaces of the film reinforce each other. This reinforcement is due to the path difference being a multiple of the wavelength of light. To find the conditions for constructive interference, we use the formula:
  • \[2n \cdot d \cdot \cos(r) = m \cdot \lambda\]
Here:
  • \(n\) is the refractive index of the film
  • \(d\) is the optical thickness of the film
  • \(\cos(r)\) considers any angle of incidence, though frequently it is normal, so \(r = 0\)
  • \(m\) is a positive integer indicating the order of interference
  • \(\lambda\) is the wavelength of the light in a vacuum
In our exercise, by setting \(m = 1\), we've sought the smallest film thickness that results in constructive interference for specific wavelengths (500 nm and 625 nm).
Wavelength
The term wavelength refers to the distance between consecutive peaks of a wave. It's crucial in determining how waves interfere with each other. For light, wavelength specifies its color, ranging from violet (shortest) to red (longest).
When dealing with thin film interference, the wavelength of light within the material is shorter due to the refractive index. The refractive index is a measure of how much the light slows down in a medium, and as such, changes the effective wavelength within the medium:
  • \[\lambda_{\text{film}} = \frac{\lambda}{n}\]
If light of multiple wavelengths reflects off a film, different colors may dominate an observed pattern. Films like soap bubbles and oil slicks are fascinating examples of this principle in action.
Optical Thickness
Optical thickness extends beyond just the physical thickness of a film; it's about the film's ability to delay light passing through. It's calculated as the product of the physical thickness and the refractive index:
  • \[d = n \cdot t\]
where \(t\) stands for the physical thickness of the film. The optical thickness determines the additional path length over which interference occurs, influencing whether constructive or destructive interference arises.
For our thin film, with a refractive index of 1.38, the minimum optical thickness derived for constructive interference of 500 nm and 625 nm wavelengths was different, but only one of these satisfies both conditions. Hence, determining the common minimum optical thickness is key to achieving simultaneous constructive interference for different wavelengths.

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Most popular questions from this chapter

Two narrow slits \(50 \mu \mathrm{m}\) apart are illuminated with light of wavelength \(500 \mathrm{nm} .\) The light shines on a screen \(1.2 \mathrm{m}\) distant. What is the angle of the \(m=2\) bright fringe? How far is this fringe from the center of the pattern?

Antireflection coatings can be used on the inner surfaces of eyeglasses to reduce the reflection of stray light into the eye, thus reducing eyestrain. a. A 90-nm-thick coating is applied to the lens. What must be the coating's index of refraction to be most effective at \(480 \mathrm{nm} ?\) Assume that the coating's index of refraction is less than that of the lens. b. If the index of refraction of the coating is \(1.38,\) what thickness should the coating be so as to be most effective at \(480 \mathrm{nm} ?\) The thinnest possible coating is best.

A 0.50-mm-wide slit is illuminated by light of wavelength \(500 \mathrm{nm} .\) What is the width of the central maximum on a screen \(2.0 \mathrm{m}\) behind the slit?

The wings of some beetles have closely spaced parallel lines of melanin, causing the wing to act as a reflection grating. Suppose sunlight shines straight onto a beetle wing. If the melanin lines on the wing are spaced \(2.0 \mu \mathrm{m}\) apart, what is the firstorder diffraction angle for green light \((\lambda=550 \mathrm{nm}) ?\)

Quality control systems have been developed to remotely measure the diameter of wires using diffraction. A wire with a stated diameter of \(170 \mu \mathrm{m}\) blocks the beam of a \(633 \mathrm{nm}\) laser, producing a diffraction pattern on a screen \(50.0 \mathrm{cm}\) distant. The width of the central maximum is measured to be \(3.77 \mathrm{mm}\). The wire should have a diameter within \(1 \%\) of the stated value. Does this wire pass the test?

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