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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 \(3.76 \times 10^{-6} \mathrm{m} ?\)

Short Answer

Expert verified
6 bright fringes can be formed on either side of the central fringe.

Step by step solution

01

Understand the problem

We are given a double slit experiment where light of wavelength \( \lambda = 625 \text{ nm} \) is used. The distance between the slits, \( d \), is \( 3.76 \times 10^{-6} \text{ m} \). We need to find how many bright fringes can appear on either side of the central maximum.
02

Use the formula for fringe maxima

In a double slit experiment, the path difference for bright fringes (constructive interference) is given by the equation: \( d \sin \theta = m \lambda \), where \( m \) is the order of the fringe (0, 1, 2, ...). We want to determine the maximum value of \( m \) for which this equation holds true.
03

Determine maximum possible value of \( m \)

The maximum order \( m \) occurs when \( \sin \theta \) is at its maximum value of 1. Thus, we need to solve \( m_{\text{max}} = \frac{d}{\lambda} \), where \( \lambda = 625 \times 10^{-9} \text{ m} \).
04

Calculate \( m_{\text{max}} \)

Substitute the values in the equation: \[ m_{\text{max}} = \frac{3.76 \times 10^{-6}}{625 \times 10^{-9}} \approx 6.016. \] Since \( m \) must be an integer, the maximum order of fringes is 6.
05

Conclude the number of bright fringes

Since the maximum order \( m \) ranging from \( -6 \) to \( 6 \) includes the central bright fringe at \( m = 0 \), there are 6 bright fringes on either side of the central fringe.

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Key Concepts

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

Wavelength
In the double slit experiment, the term "wavelength" represents the distance between consecutive peaks of a wave. For light waves, this determines the color we perceive and plays a crucial role in phenomena such as interference.
The double slit experiment uses the wavelength of light to create patterns of bright and dark fringes. In this specific problem, the wavelength is given as 625 nm (nanometers), which is equivalent to 625 x 10^-9 meters. This wavelength defines how the light waves overlap when passing through the slits.
Understanding wavelength is fundamental, as it helps us comprehend how different colors of light spread out differently when applied in the double slit setup. Moreover, it impacts the spacing of the fringes we observe. Shorter wavelengths result in fringes that are spaced closer together, while longer wavelengths spread the fringes further apart.
Bright Fringes
In the context of the double-slit experiment, bright fringes refer to the visible lines or bands of light that appear on a screen when light passes through two slits and interferes constructively. These fringes form due to constructive interference where the waves from the two slits meet in phase and reinforce each other.
The position and number of bright fringes depend heavily on the wavelength of the light and the separation between the slits. For bright fringes to occur, the path difference between the two waves must be an integer multiple of the wavelength (i.e., 0, λ, 2λ,...). This ensures that the peaks of the waves overlap precisely, producing maximum intensity at those points.
Bright fringes are significant because they serve as a visual demonstration of wave interference, allowing us to calculate variables such as the wavelength or the slit separation when certain parameters are known.
Path Difference
Path difference plays a crucial role in determining when bright fringes occur during the double slit experiment. It refers to the difference in the distances traveled by the two waves from each slit to a point on the screen.
For bright fringes, the path difference must equal an integer multiple of the wavelength. Mathematically, this is expressed as \( d \, \sin \theta = m \lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle relative to the incident light, \( \lambda \) is the wavelength, and \( m \) represents the order number of the fringe.
When the path difference fulfills this condition, it leads to constructive interference at that point, resulting in a bright fringe on the screen. This makes the path difference vital for predicting where bright long lines will appear when two waves meet after passing through the slits. By manipulating this formula, we can find out the number of fringes that can be formed under given conditions.

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Most popular questions from this chapter

In Young's experiment a mixture of orange light \((611 \mathrm{nm})\) and blue light \((471 \mathrm{nm})\) shines on the double slit. The centers of the first- order bright blue fringes lie at the outer edges of a screen that is located \(0.500 \mathrm{m}\) away from the slits. However, the first-order bright orange fringes fall off the screen. By how much and in which direction (toward or away from the slits) should the screen be moved so that the centers of the first-order bright orange fringes will just appear on the screen? It may be assumed that \(\theta\) is small, so that \(\sin \theta \approx \tan \theta\).

A diffraction grating is \(1.50 \mathrm{cm}\) wide and contains 2400 lines. When used with light of a certain wavelength, a third-order maximum is formed at an angle of \(18.0^{\circ} .\) What is the wavelength (in \(\mathrm{nm}\) )?

The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{mm}\) apart at a distance of \(3.0 \mathrm{mm}\) from the grating. Estimate the spacing between the slits of the grating.

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?

A soap film \((n=1.33)\) is \(465 \mathrm{nm}\) thick and lies on a glass plate \((n=1.52) .\) Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through the air and strikes the film perpendicularly. For which wavelength(s) in this range does destructive interference cause the film to look dark in reflected light?

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