Problem 39 - University Physics with Modern Physics

Question

Two flat plates of glass with parallel faces are on a table, one plate on the other. Each plate is 11.0 cm long and has a refractive index of 1.55. A very thin sheet of metal foil is inserted under the end of the upper plate to raise it slightly at that end, in a manner similar to that discussed in Example 35.4. When you view the glass plates from above with reflected white light, you observe that, at 1.15 mm from the line where the sheets are in contact, the violet light of wavelength 400.0 nm is enhanced in this reflected light, but no visible light is enhanced closer to the line of contact. (a) How far from the line of contact will green light (of wavelength 550.0 nm) and orange light (of wavelength 600.0 nm) first be enhanced? (b) How far from the line of contact will the violet, green, and orange light again be enhanced in the reflected light? (c) How thick is the metal foil holding the ends of the plates apart?

Step-by-Step Solution

Answer

Green light enhances at 1.58 mm, orange at 1.725 mm. Violet, green, and orange enhance again at 2.3 mm, 3.16 mm, and 3.45 mm, respectively. The foil thickness is 0.129 mm.

Step 1: Understand the problem

We want to find the positions where certain colors of light are enhanced due to constructive interference, given that the distance for violet light is known.

Step 2: Use the interference condition

In thin film interference, constructive interference occurs when the path difference \(2nt = m \lambda\), where \(n\) is the refractive index, \(t\) is the thickness of the air gap, \(m\) is an integer, and \(\lambda\) is the wavelength of light.

Step 3: Calculate for green light

For green light \(\lambda = 550.0\, \text{nm}\), using the condition \(2nt = m \lambda\), we place \(m=1\) since we're looking for the first enhancement. Given the refractive index, we can relate the air gap thickness \(t\) to the distance \(x\) from the contact line by geometry, using the relation \(t = x \cdot \tan\theta\). Since we are not given \(\theta\), we compare ratios between wavelengths: \(x_{green} = x_{violet} \cdot \frac{\lambda_{green}}{\lambda_{violet}}\).

Step 4: Solve for green light distance

Given \(x_{violet} = 1.15\, \text{mm}\) and using the method from Step 3, \(x_{green} = 1.15\, \text{mm} \cdot \frac{550\, \text{nm}}{400\, \text{nm}} = 1.58\, \text{mm}\).

Step 5: Calculate for orange light

Apply the same method as in Step 3 for orange light \(\lambda = 600.0\, \text{nm}\), using \(x_{orange} = x_{violet} \cdot \frac{\lambda_{orange}}{\lambda_{violet}}\).

Step 6: Solve for orange light distance

Similarly, \(x_{orange} = 1.15\, \text{mm} \cdot \frac{600\, \text{nm}}{400\, \text{nm}} = 1.725\, \text{mm}\).

Step 7: Repeat enhancement calculation

Enhancement happens again at the next integer \(m\). For green and orange light, use \(m=2\), \(x_{green, second} = 2 \times 1.58 \; \text{mm} = 3.16 \; \text{mm}\) and \(x_{orange, second} = 2 \times 1.725 \; \text{mm} = 3.45 \; \text{mm}\). For violet light, \(x_{violet, second} = 2 \times 1.15 \; \text{mm} = 2.3 \; \text{mm}\).

Step 8: Determine foil thickness

The foil thickness \(t\) can be related to the initial enhancement distance with \(\lambda = 400.0\, \text{nm}\). Using the relation \(t = m\frac{\lambda}{2n}\), with \(m=0\) for the first visible line due overlap, solve \(t = 1.15\times 10^{-3} \; m = m \cdot \frac{400\times 10^{-9} \; ext{m}}{2 \times 1.55}\). Calculate \(t = 0.129 \; ext{mm}\).

Key Concepts

Constructive Interference

Constructive interference occurs when two or more waves overlap, and their amplitudes reinforce each other, creating a wave of greater amplitude. In thin film interference, such as seen with soap bubbles or oil slicks, this phenomenon is observed thanks to the path difference between light waves reflected off the top and bottom surfaces of a film.
The key formula for understanding this is the constructive interference condition:

• The path difference equals \(2nt = m \lambda\),

where \(n\) is the refractive index of the film, \(t\) is the film's thickness, \(m\) is the order of the interference (an integer), and \(\lambda\) is the wavelength of the light.
When these conditions are met, the light waves are "in phase," and their energies add up, making colors bright and visible. This is why certain distances from where the plates are in contact allow for the enhancement of violet, green, or orange light through constructive interference.

Refractive Index

The refractive index is a measure of how much light slows down as it travels through a medium compared to its speed in a vacuum. It is represented by the letter \(n\). In the given exercise, the refractive index of the glass plates is 1.55, indicating that light travels 1.55 times slower in glass than in vacuum.
The refractive index plays a crucial role in determining the phase shift of light as it moves between different media.

• In the formula \(2nt = m\lambda\), \(n\) modifies the thickness of the film \(t\) in the optics calculations for interference.
• A higher refractive index implies a greater phase change and can influence where constructive or destructive interference occurs.

Understanding refractive indices helps predict how light behaves in various mediums, which is essential for our observations of thin film interference.

Wavelength of Light

The wavelength of light \(\lambda\) is the distance between consecutive crests of a wave. In the context of thin film interference, different wavelengths correspond to different colors of light. For instance, violet at 400 nm, green at 550 nm, and orange at 600 nm.
This parameter is vital, as the wavelength determines which specific colors will be enhanced or diminished due to interference.

• Shorter wavelengths, like violet, have higher energy and, often, more pronounced interference effects due to their interactions with thin films.
• Each color will have a unique point where it experiences constructive interference when the equation \(2nt = m\lambda\) is satisfied.

Thus, understanding the wavelength of light helps illustrate why different colors are observed at different positions or thickness levels in the thin film experiment.