Problem 61 - University Physics with Modern Physics

Question

The index of refraction of a glass rod is 1.48 at \(T=20.0^{\circ} \mathrm{C}\) and varies linearly with temperature, with a coefficient of \(2.50 \times 10^{-5} / \mathrm{C}^{\circ} .\) The coefficient of linear expansion of the glass is \(5.00 \times 10^{-6} / \mathrm{C}^{\circ} .\) At \(20.0^{\circ} \mathrm{C}\) the length of the rod is 3.00 \(\mathrm{cm} . \mathrm{A}\) Michelson interferometer has this glass rod in one arm, and the rod is being heated so that its temperature increases at a rate of 5.00 \(\mathrm{C}^{\circ} / \mathrm{min}\) . The light source has wavelength \(\lambda=589 \mathrm{nm},\) and the rod initially is at \(T=20.0^{\circ} \mathrm{C}\) . How many fringes cross the field of view each minute?

Step-by-Step Solution

Answer

Approximately 16.5 fringes cross the field of view each minute.

Step 1: Understand the Problem

We need to calculate how many fringes of a Michelson interferometer will cross the field of view in a minute due to the change in temperature of a glass rod. The change in temperature affects both the refractive index and the length of the rod.

Step 2: Calculate Change in Refractive Index Per Minute

The refractive index change rate with temperature is given by the coefficient of variation of refractive index: \[ \Delta n = \left(2.50 \times 10^{-5} \right) \times 5.00 = 1.25 \times 10^{-4} \, \text{per minute} \]

Step 3: Calculate Change in Length Per Minute

The length of the rod changes due to thermal expansion. The change in length is calculated by: \[ \Delta L = L_0 \times \alpha \times \Delta T = 3.00 \, \text{cm} \times 5.00 \times 10^{-6} \, / \mathrm{C}^{\circ} \times 5.00 = 7.50 \times 10^{-5} \, \text{cm/min} \]

Step 4: Calculate Optical Path Difference

The optical path difference (OPD) change in the interferometer is due to both changes in length and refractive index. It is:\[ \Delta(\text{OPD}) = 2\left(n \Delta L + L_0 \Delta n \right) \]

Step 5: Substitute Values and Simplify

Substitute the calculations:1. \( \Delta L = 7.50 \times 10^{-5} \text{ cm/min} \)2. \( L_0 = 3.00 \text{ cm} \)3. \( \Delta n = 1.25 \times 10^{-4} \text{ per minute} \)4. \( n = 1.48 \)\[ \Delta(\text{OPD}) = 2 \left(1.48 \times 7.50 \times 10^{-5} + 3.00 \times 1.25 \times 10^{-4} \right) \]

Step 6: Calculate the Number of Fringes

Calculate how many wavelengths fit into the change in optical path difference:\[ \Delta(\text{OPD}) = 2 \left(1.11 \times 10^{-4} + 3.75 \times 10^{-4} \right) = 9.72 \times 10^{-4} \text{ cm/min} \]The number of fringes is:\[ \frac{\Delta(\text{OPD})}{\lambda} = \frac{9.72 \times 10^{-4}}{589 \times 10^{-7}} \approx 16.5 \text{ fringes/min} \]

Step 7: Conclusion

Approximately 16.5 fringes cross the field of view each minute.

Key Concepts

Refractive Index

The refractive index is a measure of how much a material can bend light. When light enters a new medium, its speed changes, resulting in refraction. The refractive index quantifies this change. For a glass rod, this property can change with temperature variations. In the given problem, we're told the refractive index at \(20^\circ C\) is 1.48, and it changes linearly with temperature. This means as the glass rod heats up, its refractive index increases or decreases by a set amount per degree Celsius, specified by a coefficient \( \left(2.50 \times 10^{-5}\right) / \mathrm{C^\circ} \).
To understand this further:

• Higher refractive index means light slows down more.
• Temperature affects this by causing the glass's physical properties to change, altering how light travels through it.

Refractive index changes play a crucial role in calculating the optical path difference, especially in systems like the Michelson interferometer, where precise measurements of light paths are crucial.

Thermal Expansion

Thermal expansion is the process by which materials change in size due to changes in temperature. In the glass rod used in the Michelson interferometer, this occurs linearly as temperature rises. The expansion is quantified by the coefficient of linear expansion, \( \alpha = 5.00 \times 10^{-6} / \mathrm{C^\circ} \).
What this means is:

• The glass rod's length increases by a small fraction with each degree increase in temperature.
• In this example, at the initial temperature of \(20^\circ C\), the rod is 3.00 cm long. As it heats up, it expands at a rate determined by the coefficient.

Understanding thermal expansion is vital in optics, especially with devices where precise lengths affect measurement outcomes. In the interferometer, this expansion contributes to the change in optical path difference.

Optical Path Difference

Optical Path Difference (OPD) is the effective path length that light appears to travel through an optical system. It takes into account both the actual physical length and the refractive index of the medium.
In the Michelson interferometer setup:

• The OPD changes due to the refractive index shift and the physical expansion of the rod when heated.
• It is key in determining interference patterns, as even small differences result in observable changes in the interference fringes.
• The OPD is calculated as \( \Delta(\text{OPD}) = 2(n \Delta L + L_0 \Delta n) \).

Understanding OPD helps explain how temperature affects light paths and is essential for determining fringe shifts in interferometry.

Fringe Counting

Fringe counting is a technique used in interferometry to measure changes with high precision. It involves counting the interference fringes produced when two light waves overlap.
In the given problem:

• Each fringe corresponds to a complete cycle of constructive and destructive interference.
• The number of fringes that shift or cross the field of view indicates an accumulated optical path difference.
• By using the formula \( \frac{\Delta(\text{OPD})}{\lambda} \), the number of fringes per minute can be calculated, showing how much the combined changes in index and length affect the path of light.

Fringe counting is essential to precisely quantify changes in optical systems like a Michelson interferometer, providing insight into phenomena like thermal changes.

Wavelength of Light

The wavelength of light, denoted by \( \lambda \), is the distance over which the wave's shape repeats. For the Michelson interferometer problem presented, the wavelength \( \lambda = 589 \mathrm{nm} \) is involved.
When considering light behavior:

• The exact wavelength determines how light will interfere and hence, form fringes.
• The wavelength affects how many fringes will be observed for a given optical path difference.
• In this problem, the wavelength is crucial in the fringe counting formula \( \frac{\Delta(\text{OPD})}{\lambda} \).

Wavelength is a fundamental property that affects many optical phenomena, crucial for understanding how light interferes in technologies such as interferometers.