91Ó°ÊÓ

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_{0}\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_{0}=496 \mathrm{nm},\) constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is \(386 \mathrm{nm}\). (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Step by step solution

01

Formula for Path Difference

First, realize that the path difference for the light which transmits through the glass and experiences two reflections is \(2nd = m\lambda\), where \(m\) is an integer (0, 1, 2, ...), \(n\) is the refractive index of the medium (in this case the glass), \(d\) is the thickness of the glass, and \(\lambda\) is the wavelength of light in the medium (which is \(\lambda / n\) when \( \lambda_{0} \) is the wavelength in vacuum). From this equation, the thickness of the glass can be found as \(d = m\lambda / (2n)\).

02

Find m for First Constructive Interference

First, use the given \(\lambda_{0} = 496 \) nm and the refractive index \(n = 1.40\). Substitute these values into the formula from step 1 to find \(d = m(496/1.40) / (2 * 1.40)\). From this formula, find the value of the integer \(m\)

03

Find m for Second Constructive Interference

When the next shorter vacuum wavelength, \(\lambda_{0} = 386 \) nm, leads to constructive interference, the integer \(m\) would decrement by 1. So, follow the procedure in step 2 but with \(m - 1\) for \(m\). So, \(d = (m - 1)(386/1.40) / (2 * 1.40)\)

04

Equating the thickness for solution

Now equate the two expressions for 'd' we've obtained in steps 2 and 3 to cancel out 'd'. After simplifying, find the value of 'm' and consequently 'd'.

05

Longest Wavelength Calculation

For the longest wavelength, we need to use the minimum possible m, which is m=1. Substitute m=1 into the formula \(d = m\lambda / (2n)\) to get \( \lambda = 2nd/m \). Substitute the known values to calculate the longest wavelength for which there is constructive interference.

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Key Concepts

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

Refractive Index

The refractive index, often symbolized by "n," is a measure of how much light bends, or refracts, as it passes through a substance. When light enters a new medium, its speed changes. This change in speed causes light to bend, and the refractive index quantifies that bending.

• A higher refractive index indicates that light slows down more and bends more when entering the medium.
• A value of "1" means light travels at the same speed as in a vacuum.

In our exercise, the glass has a refractive index of 1.40. This means that light travels slower in the glass than in the vacuum or the gas, which have refractive indices close to unity. The refractive index is crucial for calculating the thickness of the glass since it affects the light's path length and thus the interference pattern observed.

Interference

Interference is a phenomenon where two waves overlap, and their effects "interfere" with each other. This can be constructive (waves adding together) or destructive (waves canceling out).

• Constructive interference occurs when waves are in phase, meaning their peaks align.
• Destructive interference happens when waves are out of phase, meaning the peak of one wave aligns with the trough of another.

In optics, interference is responsible for the colorful patterns seen in thin films, like soap bubbles. In our exercise, the light waves reflected from the top and bottom surfaces of the glass interfere. When certain wavelengths lead to constructive interference, bright bands are visible. By observing these patterns, the thickness of the glass can be determined.

Coherent Light

Coherent light is light in which waves maintain a constant phase relationship. This property is essential for interference to occur robustly and predictably.

• Lasers are common sources of coherent light.
• Coherence ensures that the waves align in a way that allows interference patterns to be observed reliably.

In the exercise, a coherent light source is used so that when light reflects from different surfaces, the interference pattern is clear and stable. Without coherency, the interference effects would blur, making it difficult to measure accurate distances or thicknesses.

Wavelength

Wavelength is the distance between two consecutive peaks of a wave. In optics, the wavelength of light determines its color. Light behaves differently as its wavelength changes when it enters different media.

• Shorter wavelengths correspond to higher energies and deeper colors (like violet and blue).
• Longer wavelengths are associated with lower energies and colors like red and yellow.

In our scenario, the constructive interference at particular wavelengths (496 nm and 386 nm) helps determine glass thickness. As light enters the glass, its wavelength decreases to \( \lambda = \frac{\lambda_{0}}{n} \). By knowing these wavelengths and using the refractive index, calculations can be done to find out the physical thickness of the glass layer needed to cause these interference conditions.