91Ó°ÊÓ

Crude Thickness Monitor. You and your team need to determine the thicknesses of two extremely thin sheets of plastic and do not have a measurement instrument capable of the job. You find a white light source with color filters (i.e., tinted glass windows that only let certain colors pass through) including red \((\lambda=660 \mathrm{nm}),\) green \((\lambda=550 \mathrm{nm}),\) and blue \((\lambda=\) \(470 \mathrm{nm}) .\) You locate two glass microscope slides, each of length \(10.0 \mathrm{cm}\) stack one on top of the other, and tape their ends together on one side. On the very edge of the side opposite the tape, you sandwich a piece of plastic between the slides to form a wedge of air between the plates. (a) When you shine blue light perpendicular to the slides, 55 bright fringes form along the full length of the slide. How thick is the plastic? (b) When you replace the first sheet with the second sheet of plastic, the fringes that appear are too closely spaced to count. You switch to red light (a longer wavelength), and count 124 fringes. How thick is the plastic? (c) How many fringes would you expect in each case, (a) and (b), if you instead used green light?

Step by step solution

01

Understand the Problem

You need to determine the thickness of two thin plastic sheets by using the interference of light fringes. This problem leverages the concept of thin film interference which results from the wedge-shaped air gap between the plates.

02

Apply the Interference Formula

For thin film interference, where the path difference results in bright fringes, we use the formula: \( 2d = m \lambda \), where \( d \) is the thickness of the air wedge, \( \lambda \) is the wavelength of light, and \( m \) is the order of the fringe (number of fringes).

03

Calculate Thickness for Blue Light (Part a)

Given: \( \lambda_{blue} = 470 \ nm \), fringes \( m = 55 \), and slide length \( L = 10 \ cm = 100 \ mm \). The path difference \( 2d = m \lambda \) gives us \( d = \frac{m \lambda}{2} \). Therefore, \( d = \frac{55 \times 470 \ nm}{2 \times 100 \ mm} \rightarrow d = \frac{25850 \ nm}{200 \ mm} = 129.25 \ nm \).

04

Calculate Thickness for Red Light (Part b)

Given: \( \lambda_{red} = 660 \ nm \) and fringes \( m = 124 \). Similarly, \( d = \frac{124 \times 660 \ nm}{2 \times 100 \ mm} = \frac{81840 \ nm}{200 \ mm} = 409.2 \ nm \).

05

Calculate Expected Fringes for Green Light (Part c) Case (a)

Given: \( \lambda_{green} = 550 \ nm \). For the first plastic sheet (\( d = 129.25 \ nm \)), using \( m = \frac{2d \times 100 \ mm}{\lambda} \): \( m = \frac{2 \times 129.25 \ nm}{550 \ nm} \times 100 \ mm = 47 \).

06

Calculate Expected Fringes for Green Light (Part c) Case (b)

For the second plastic sheet (\( d = 409.2 \ nm \)), using the same formula: \( m = \frac{2 \times 409.2 \ nm}{550 \ nm} \times 100 \ mm = 149 \).

07

Verify and Conclude

Check calculations and ensure logical consistency in answers. Conclude that the first plastic thickness is approximately 129.25 nm and the second is approximately 409.2 nm.

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Key Concepts

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

Interference of Light

Interference of light occurs when two or more light waves overlap, causing them to either reinforce or cancel each other. When light waves are reflected or transmitted through different mediums, they can create a pattern of alternating bright and dark stripes known as interference fringes. These fringes are a direct result of constructive and destructive interference.

In this particular problem, the thin plastic sheets cause a wedge-shaped air gap between two glass slides. This setup results in thin film interference. When light passes through this wedge, it reflects on both the top and bottom surfaces of the wedge. The path difference between these two reflected light waves leads to interference. Based on the constructive interference, bright fringes appear, representing areas where the light waves are reinforcing each other.

Fringe Formation

The formation of fringes is a visual representation of the interference of light. It results from the variation in the thickness of the air wedge between the glass slides with a plastic sheet sandwiched in between. The alternating bright and dark bands that are observed are known as fringes.

These fringes become visible because light waves travelling slightly different paths due to the wedge shape, either reinforce each other (bright fringes) or cancel out (dark fringes). The edges of the wedge cause a varying path length for light waves, resulting in different fringe patterns when observed from different angles or with different wavelengths. The number and type of fringes (bright or dark) are influenced by the wavelength of the light and the thickness of the wedge.

• Bright fringes: Arise from constructive interference where waves combine.
• Dark fringes: Result from destructive interference where waves cancel out.

Wavelength Calculations

Calculating the values associated with interference patterns frequently involves the formula for constructive interference: \( 2d = m \lambda \). Here, \( d \) is the thickness of the thin film (air wedge), \( \lambda \) is the wavelength of the light used, and \( m \) is the fringe order or number of fringes observed.

In the given exercise:

• For blue light: with \( \lambda_{blue} = 470 \, nm \) and \( m = 55 \), thickness \( d \) was calculated to be approximately 129.25 nm.
• For red light: with \( \lambda_{red} = 660 \, nm \) and \( m = 124 \), \( d \) was about 409.2 nm.
• For green light: individuals would need to substitute \( \lambda_{green} = 550 \, nm \) into the formula to determine the expected fringe number from the known thickness.

These calculations are essential for estimating the unknown thickness of thin films by exploiting how different wavelengths lead to different interference patterns.