Question

One round face of a 3.25-m, solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 mm apart, are made in the center of the black face. When laser light of wavelength 632.8 nm shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 mm wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

Short answer

The index of refraction of the plastic is approximately 1.49.

Step-by-step solution

Identify Known Values

First, we note the known values from the problem. The distance between the slits, or the slit separation, is $d=0.225\text{mm}$. This needs to be converted to meters: $d=0.225\times {10}^{-3}\text{m}$. The wavelength of the laser light $\lambda =632.8\text{nm}=632.8\times {10}^{-9}\text{m}$. The width of the central bright fringe $w=5.82\text{mm}=5.82\times {10}^{-3}\text{m}$.

Understand the Relationship

We know that fringe width $w$ in the double-slit interference pattern is given by the formula $w=\frac{2\lambda L}{d}$, where $L$ is the distance between the slits and the screen (opposite face of the cylinder). Since this pattern forms inside the plastic cylinder, $L$ is related to the cylinder's radius $R=\frac{3.25}{2}\approx 1.625\text{m}$.

Adjust for Medium's Index of Refraction

Since the light travels through the plastic, we must account for the index of refraction $n$. The apparent wavelength in a medium is given by ${\lambda }^{\prime }=\frac{\lambda }{n}$. Thus, the width of the fringe considering the medium becomes $w=\frac{2{\lambda }^{\prime }L}{d}=\frac{2\lambda L}{nd}$.

Solve for the Index of Refraction

By rearranging the formula for $w$, we get $n=\frac{2\lambda L}{wd}$. Substituting the known values: $n=\frac{2\times 632.8\times {10}^{-9}\text{m}\times 1.625\text{m}}{5.82\times {10}^{-3}\text{m}\times 0.225\times {10}^{-3}\text{m}}$. Calculating this gives $n\approx 1.49$.

Key concepts

Double-Slit Interference

Double-slit interference is a fascinating phenomenon of optical physics that happens when light waves pass through two closely spaced slits and interfere with each other. This forms a pattern of alternating bright and dark bands, known as fringes, on a screen or surface.

The bright fringes occur where the light waves meet and amplify each other (constructive interference), while the dark fringes occur where the waves cancel each other out (destructive interference).

• The distance between two adjacent bright or dark fringes is called the "fringe width."
• The central bright fringe is typically the most intense and is located exactly opposite the slits.

The formula for fringe width, commonly given by $w=\frac{2\lambda L}{d}$, helps us identify how the light's wavelength, the slit separation $d$, and the radius of the medium (cylinder, in this case) $L$, influence this pattern.

Understanding double-slit interference is crucial for designing experiments and technologies that depend on precise measurements of light's behavior.

Index of Refraction

The index of refraction $n$ is a fundamental property of materials, crucial for understanding how light behaves as it passes through them. It indicates how much light slows down in a material compared to its speed in a vacuum.

Mathematically, the index of refraction is defined as the ratio of the speed of light in a vacuum $c$ to the speed of light in the medium $v$, expressed as $n=\frac{c}{v}$. A higher index means light travels more slowly through the medium. This alteration in speed occurs because light waves are bent, or refracted, which is central to navigating optical designs such as lenses.

• When light enters a medium with an index of refraction greater than one, its wavelength decreases.
• This adjustment impacts how interference patterns like double-slit interference are formed.

Utilizing the formula $n=\frac{2\lambda L}{wd}$, where adjustments consider both the intrinsic and apparent properties of light in different media, we can determine the index of refraction from observed patterns, as demonstrated in the original problem.

Light Wavelength

Light wavelength is the distance between successive peaks of a light wave and is a crucial determinant of its behavior and properties.

Measured in nanometers (nm) for visible light, this is what gives light its color — longer wavelengths appear redder, whereas shorter wavelengths are bluer. In the context of the double-slit interference experiment, the wavelength $\lambda$ plays a significant role in determining the fringe patterns formed.

• Just like sound waves, changes in the medium's density impact light's wavelength by altering its speed.
• This relationship is adjusted by the material's index of refraction, affecting how the light wave behaves.

Thus, laser light of a specific known wavelength, such as the 632.8 nm laser used in this problem, lets us predict and interpret interference effects.

These predictable effects enable practical applications like determining material indexes of refraction, as more energetic (shorter wavelength) or less energetic (longer wavelength) light interacts differently based on the medium it passes through.