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Chapter 27: Problem 5

In a Young's double-slit experiment, the seventh dark fringe is located \(0.025 \mathrm{m}\) to the side of the central bright fringe on a flat screen, which is \(1.1 \mathrm{m}\) away from the slits. The separation between the slits is \(1.4 \times 10^{-4} \mathrm{m}\) What is the wavelength of the light being used?

Short Answer

Expert verified
The wavelength of the light is approximately 424 nm.

Step by step solution

01

Understand the Concept

In a Young's double-slit experiment, the dark and bright fringes are caused by the interference of light waves. The dark fringes occur where the path difference between the two waves is a half-integer multiple of the wavelength.
02

Use the Formula for Dark Fringes

The path difference for dark fringes is given by the formula \(d \sin \theta = (m + 0.5)\lambda\), where \(d\) is the distance between the slits, \(\theta\) is the angle of the dark fringe, \(\lambda\) is the wavelength of light, and \(m\) is the order number of the dark fringe.
03

Relate Distance and Angle

For small angles, \(\sin \theta \approx \tan \theta\). The angle \(\theta\) can be found using \(\tan \theta = \frac{y}{L}\), where \(y\) is the distance from the central fringe (0.025 m) and \(L\) is the distance to the screen (1.1 m).
04

Calculate the Angle Theta

Calculate \(\tan \theta\) for the seventh dark fringe as follows:\[ \tan \theta \approx \frac{0.025}{1.1} \approx 0.0227 \]Since \(\sin \theta \approx \tan \theta\), we use \(\sin \theta \approx 0.0227\).
05

Solve for the Wavelength

Use the equation for the dark fringe: \[1.4 \times 10^{-4} \times 0.0227 = (7 + 0.5)\lambda\]Solve for \(\lambda\):\[\lambda = \frac{1.4 \times 10^{-4} \times 0.0227}{7.5} \approx 4.24 \times 10^{-7} \text{ m}\] or \(424 \text{ nm}\).

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Key Concepts

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

Understanding Interference of Light
Interference is a fundamental concept in wave physics where two waves superimpose to form a resultant wave of greater, lower, or the same amplitude. It can be described as constructive or destructive interference when it results in increased or decreased light intensity, respectively. In Young's double-slit experiment, interference of light occurs when light passes through two closely spaced slits and overlaps.
- **Constructive Interference**: This occurs when the path difference between the waves from the two slits is an integer multiple of the wavelength, leading to bright fringes on the screen. - **Destructive Interference**: Here, the waves from the two slits are out of phase by half a wavelength, leading to cancellation and dark fringes on the screen. Young鈥檚 experiment beautifully demonstrates this by producing an interference pattern consisting of alternating bright and dark bands or fringes, evidencing the wave nature of light.
Understanding Dark Fringes in Interference
Dark fringes are an integral part of the interference pattern seen in Young's experiment. They manifest where destructive interference occurs, involving phase differences that cancel out the light waves.
The order or number of a specific dark fringe, denoted as "m," helps identify its position relative to the central fringe. - If the path difference from the two slits is \( (m + 0.5)\lambda \) (where \(m\) is an integer and \(\lambda\) the wavelength), a dark fringe will form.- The first dark fringe (m=0) appears closest to the center, and the order number increases as we move further from the center.Calculating the position of these dark fringes involves understanding the geometry of the experiment, where small angle approximations (\( \sin \theta \approx \tan \theta \)) are often employed for ease of calculation. This simplifies the process of relating a fringe's position to its interference conditions.
Calculating Wavelength from Fringe Patterns
One fascinating outcome of Young's double-slit experiment is the ability to calculate the wavelength of the light used from the interference pattern.
By understanding the relationship between the slit separation, screen distance, and fringe position, it's possible to infer the light's wavelength using the formula for dark fringes:\[d \sin \theta = (m + 0.5)\lambda\]**Steps to Calculate Wavelength**:- **Identify Variables**: Know the slit separation \(d\), the distance to the screen \(L\), and the distance from the central fringe to the dark fringe \(y\).- **Calculate \(\theta\)**: Use \( \tan \theta = \frac{y}{L} \approx \sin \theta \) for small angles to find \( \theta\).- **Solve for Wavelength \(\lambda\)**: Rearrange and solve the dark fringe equation to find \(\lambda\).This calculated wavelength reveals the precise characteristics of the light source, bridging theoretical and experimental physics with elegant simplicity.

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Most popular questions from this chapter

An inkjet printer uses tiny dots of red, green, and blue ink to produce an image. Assume that the dot separation on the printed page is the same for all colors. At normal viewing distances, the eye does not resolve the individual dots, regardless of color, so that the image has a normal look. The wavelengths for red, green, and blue are \(\lambda_{\text {red }}=660 \mathrm{nm}, \lambda_{\text {green }}=550 \mathrm{nm},\) and \(\lambda_{\text {blue }}=470 \mathrm{nm} .\) The diameter of the pupil through which light enters the eye is \(2.0 \mathrm{mm}\). For a viewing distance of \(0.40 \mathrm{m},\) what is the maximum allowable dot separation?

Two in-phase sources of waves are separated by a distance of \(4.00 \mathrm{m}\). These sources produce identical waves that have a wavelength of \(5.00 \mathrm{m}\). On the line between them, there are two places at which the same type of interference occurs. (a) Is it constructive or destructive interference, and (b) where are the places located?

A tank of gasoline \((n=1.40)\) is open to the air \((n=1.00) .\) A thin film of liquid floats on the gasoline and has a refractive index that is between 1.00 and \(1.40 .\) Light that has a wavelength of \(625 \mathrm{nm}\) (in vacuum) shines perpendicularly down through the air onto this film, and in this light the film looks bright due to constructive interference. The thickness of the film is \(242 \mathrm{nm}\) and is the minimum nonzero thickness for which constructive interference can occur. What is the refractive index of the film?

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength \(625 \mathrm{nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{m} ?\)

A film of oil lies on wet pavement. The refractive index of the oil exceeds that of the water. The film has the minimum nonzero thickness such that it appears dark due to destructive interference when viewed in red light (wavelength \(=640.0 \mathrm{nm}\) in vacuum). Assuming that the visible spectrum extends from 380 to \(750 \mathrm{nm},\) for which visible wavelength(s) in vacuum will the film appear bright due to constructive interference?
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