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Monochromatic light of wavelength \(441 \mathrm{~nm}\) is incident on a narrow slit. On a screen \(2.00 \mathrm{~m}\) away, the distance between the second diffraction minimum and the central maximum is \(1.50 \mathrm{~cm}\). (a) Calculate the angle of diffraction \(\theta\) of the second minimum. (b) Find the width of the slit.

Short Answer

Expert verified
(a) \( \theta \approx 0.430 \) degrees; (b) Slit width \( a \approx 1.176 \times 10^{-4} \ \text{m} \).

Step by step solution

01

Understand the Diffraction Minimum Concept

For a single slit diffraction pattern, the minima occur at angles \( \theta \) where \( a \sin \theta = m \lambda \). Here, \( a \) is the slit width, \( m \) is the order of the minimum, \( \lambda \) is the wavelength, and \( \theta \) is the angle of diffraction. For the second minimum, \( m = 2 \).
02

Relate Angle to Linear Distance

The angle \( \theta \) and the linear distance \( y \) on the screen are related by \( y = L \tan \theta \), where \( L \) is the distance from the slit to the screen. For small angles, \( \tan \theta \approx \sin \theta \). This gives \( y = L \sin \theta \).
03

Calculate the Angle of Diffraction \(\theta\)

Given \( y = 1.50 \ \text{cm} = 0.0150 \ \text{m} \) and \( L = 2.00 \ \text{m} \), we have \( \sin \theta = \frac{y}{L} = \frac{0.0150}{2.00} = 0.0075 \). Therefore, \( \theta = \arcsin(0.0075) \approx 0.430 \) degrees.
04

Use the Diffraction Equation to Find the Slit Width

Using the equation for the minima \( a \sin \theta = m \lambda \), for the second minimum \( m = 2 \), and \( \lambda = 441 \ \text{nm} = 441 \times 10^{-9} \ \text{m} \), we solve for \( a \): \( a = \frac{m \lambda}{\sin \theta} = \frac{2 \times 441 \times 10^{-9}}{0.0075} \approx 1.176 \times 10^{-4} \ \text{m} \).

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Key Concepts

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

Monochromatic Light
Monochromatic light refers to a light source that emits light of a single wavelength. This means that all the light waves are coherent, which helps in creating consistent and clear patterns when they interfere with each other. For instance, when monochromatic light passes through a narrow slit, it encounters diffraction, creating distinct patterns on a screen. A common source of monochromatic light used in experiments is lasers or specific types of LEDs. With a known wavelength of the light, like the given 441 nm in this exercise, calculations can be made regarding diffraction patterns, especially in controlled environments. Monochromatic light is crucial in experiments where precise measurements are required, as it reduces the complexity arising from multiple wavelengths.
Slit Width
Slit width is a critical factor in determining the diffraction pattern when light passes through an opening. It is defined as the physical measurement from one side of the slit to the other, usually in meters or fractions thereof. The width of the slit, denoted as \( a \) in equations, plays a significant role in the formation of diffraction patterns. A narrower slit causes more significant spreading or bending of light waves, leading to more pronounced diffraction effects. This is evident by observing a wider spacing between diffraction minima and maxima on a screen. In our exercise, the slit width is calculated using the diffraction equation \( a \sin \theta = m \lambda \). The calculated slit width helps us understand how tightly or loosely the light is being diffracted.
Diffraction Pattern
A diffraction pattern is the series of light and dark bands seen on a screen resulting from the bending of light waves when they encounter an obstacle, like a slit. These patterns are the result of constructive and destructive interference of the light waves. When light waves overlap constructively, bright bands appear, whereas dark bands indicate destructive interference. For monochromatic light passing through a single slit, this pattern consists of a central maximum flanked by multiple minima and secondary maxima. The size and spacing of these fringes depend on the wavelength of the light and the width of the slit. Understanding the diffraction pattern allows scientists to deduce properties like slit width or the wavelength of light used in the experiment. It provides insights into the wave nature of light by showcasing how different paths and overlaps of light influence what we observe.
Angle of Diffraction
The angle of diffraction, represented by \( \theta \), is the angle at which light waves emerge after passing through a slit, causing them to spread out rather than continuing in a straight line. This angle depends on the wavelength of the light and the slit width. In single-slit diffraction, the angle is related to positions of minima (dark fringes) on the resulting diffraction pattern.
  • The formula \( a \sin \theta = m \lambda \) connects angle with wavelength and slit width, where \( m \) is the order of the minima.

  • For the second minimum (\( m = 2 \)), knowing the distance to the screen and the distance from the center to the second dark fringe allows one to calculate \( \theta \).

  • This angle is calculated as \( \theta = \arcsin \left( \frac{0.0150 \text{ m}}{2.00 \text{ m}} \right) \approx 0.430 \) degrees.
Understanding the angle of diffraction helps us visualize how light behaves under constraints and provides a quantitative way to predict how light will propagate after passing through obstacles.

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Most popular questions from this chapter

Estimate the linear separation of two objects on Mars that can just be resolved under ideal conditions by an observer on Earth (a) using the naked eye and (b) using the 200 in. \((=5.1 \mathrm{~m})\) Mount Palomar telescope. Use the following data: distance to Mars \(=8.0 \times 10^{7} \mathrm{~km}\), diameter of pupil \(=5.0 \mathrm{~mm}\), wavelength of light \(=550 \mathrm{~nm} .\)

(a) What is the angular separation of two stars if their images are barely resolved by the Thaw refracting telescope at the Allegheny Observatory in Pittsburgh? The lens diameter is \(76 \mathrm{~cm}\) and its focal length is \(14 \mathrm{~m}\). Assume \(\lambda=550 \mathrm{~nm}\). (b) Find the distance between these barely resolved stars if each of them is 10 light-years distant from Earth. (c) For the image of a single star in this telescope, find the diameter of the first dark ring in the diffraction pattern, as measured on a photographic plate placed at the focal plane of the telescope lens. Assume that the structure of the image is associated entirely with diffraction at the lens aperture and not with lens "errors."

If you look at something \(40 \mathrm{~m}\) from you, what is the smallest length (perpendicular to your line of sight) that you can resolve, according to Rayleigh's criterion? Assume the pupil of your eye has a diameter of \(4.00 \mathrm{~mm}\), and use \(500 \mathrm{~nm}\) as the wavelength of the light reaching you.

The pupil of a person's eye has a diameter of \(5.00 \mathrm{~mm}\). According to Rayleigh's criterion, what distance apart must two small objects be if their images are just barely resolved when they are \(250 \mathrm{~mm}\) from the eye? Assume they are illuminated with light of wavelength \(500 \mathrm{~nm}\).

Two yellow flowers are separated by \(60 \mathrm{~cm}\) along a line perpendicular to your line of sight to the flowers. How far are you from the flowers when they are at the limit of resolution according to the Rayleigh criterion? Assume the light from the flowers has a single wavelength of \(550 \mathrm{~nm}\) and that your pupil has a diamcter of \(5.5 \mathrm{~mm}\).

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