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The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen and the slit. Find the ratio \(\lambda / W\) of the wavelength \(\lambda\) of the light to the width \(W\) of the slit.

Short Answer

Expert verified
The ratio \( \lambda / W \) is \( 1/2 \).

Step by step solution

01

Understanding the problem

We are asked to find the ratio \( \lambda / W \) given that the central bright fringe width in a single-slit diffraction pattern equals the distance to the screen. This suggests that the geometry of the problem relates the slit width \( W \), the wavelength \( \lambda \), and the distance to the screen \( L \).
02

Concept of single-slit diffraction

In single-slit diffraction, the angular position of the minima is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle from the central maximum, \( m \) is the order of the minimum, and \( \lambda \) is the wavelength. The central bright fringe width is determined by the distance between the first minima on either side of the central maximum.
03

Relation to central bright fringe width

The central bright fringe width \( W' \) can be approximated as \( W' = 2L \tan \theta \) where \( \theta \) is the angle to the first minimum. For very small angles, \( \tan \theta \approx \sin \theta \).
04

Finding the first minimum

For the first minimum, \( m = 1 \), so the condition becomes \( a \sin \theta = \lambda \), and therefore \( \sin \theta = \lambda / a \).
05

Using given condition of width

We are given that the central bright fringe width equals the distance \( L \) to the screen: \( W' = L \). Since \( W' = 2L \sin \theta \) and \( \sin \theta = \lambda / W \), we write \( W' = 2L \cdot \lambda / W = L \).
06

Solving for \( \lambda / W \)

This simplifies to \( 2 \lambda = W \), therefore \( \lambda / W = 1/2 \).

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Key Concepts

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

Wavelength
The concept of wavelength is central to understanding light's behavior in various phenomena like diffraction. Wavelength, denoted by \( \lambda \), is the distance between successive crests of a wave. In the context of light, it determines the color we perceive, but it also plays a crucial role in diffraction patterns.

In single-slit diffraction, the wavelength influences how light bends around obstacles. The bending results in patterns of light and dark fringes, known as diffraction patterns. These patterns are visible when light passes through a narrow slit. The dimensions of these patterns depend heavily on the light's wavelength.

It's important to grasp how wavelength affects diffraction because it provides insights into the wave nature of light. You can think of frequency and wavelength as inversely proportional. A longer wavelength means lower frequency and vice versa. Using this understanding can help predict how different wavelengths will behave in various diffraction scenarios.
Central Bright Fringe
In single-slit diffraction, the central bright fringe is the most prominent feature of the diffraction pattern. It is the bright band that appears directly opposite the slit. This central fringe is larger in width compared to other fringes.

The width of the central bright fringe is determined by the position of the first minima on either side. These minima are the points where the light's intensity reaches zero.

To understand the central bright fringe, let’s consider its formation:
  • The width of this fringe is represented as \( W' \).
  • It is calculated based on the angle \( \theta \) to the first minimum (dark fringe).
  • The formula \( W' = 2L \tan \theta \) helps in determining the width, where \( L \) is the distance to the screen.
This concept is crucial as it connects the central bright feature of the diffraction pattern to the physical parameters like wavelength and slit width, demonstrating the practical aspects of light behavior.
Minima
In the realm of diffraction, minima are points where the light's intensity falls to zero, creating dark fringes in the diffraction pattern. These dark bands alternate with bright fringes, adding a characteristic look to the diffraction patterns.

Minima occur due to destructive interference, where waves cancel each other out. In a single-slit diffraction setup, the condition for minima is given by the formula:
  • \( a \sin \theta = m \lambda \)
Here, \( a \) is the slit width, \( \theta \) is the angle to the minimum, \( m \) is the order of the minima (1, 2, 3,...), and \( \lambda \) is the wavelength. This formula tells us that the angle at which minima occur depends directly on the order \( m \) and inversely on the slit width.

Understanding minima is essential as they define the boundaries of the diffraction pattern fringes. In practice, identifying these points helps in calculating important features like the central bright fringe and evaluating the effects of variation in wavelength and slit dimensions.

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Most popular questions from this chapter

A spotlight sends red light (wavelength \(=694.3 \mathrm{nm}\) ) to the moon. At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{m}\) away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about \(0.20 \mathrm{m},\) and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon.

In a single-slit diffraction pattern on a flat screen, the central bright fringe is \(1.2 \mathrm{cm}\) wide when the slit width is \(3.2 \times 10^{-5} \mathrm{m} .\) When the slit is replaced by a second slit, the wavelength of the light and the distance to the screen remaining unchanged, the central bright fringe broadens to a width of \(1.9 \mathrm{cm} .\) What is the width of the second slit? It may be assumed that \(\theta\) is so small that \(\sin \theta \approx \tan \theta .\)

In a Young's double-slit experiment, two rays of monochromatic light emerge from the slits and meet at a point on a distant screen, as in Figure 27.6a. The point on the screen where these two rays meet is the eighth-order bright fringe. The difference in the distances that the two rays travel is \(4.57 \times\) \(10^{-6} \mathrm{m} .\) What is the wavelength (in \(\mathrm{nm}\) ) of the monochromatic light?

Violet light (wavelength \(=410 \mathrm{nm}\) ) and red light (wavelength \(=\) \(660 \mathrm{nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the thirdorder maximum. (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

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?

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