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Chapter 27: Problem 4

The dark fringe for \(m=0\) in a Young's double-slit experiment is located at an angle of \(\theta=15^{\circ} .\) What is the angle that locates the dark fringe for \(m=1 ?\)

Short Answer

Expert verified
No valid angle exists for \(m=1\) within expected parameters or \( \theta \approx 3 \times 15^{\circ}\). Consider highlights or bar in standards.

Step by step solution

01

Understanding the Problem

In a Young's double-slit experiment, dark fringes occur at angles where the path difference between light from the two slits is a half-integer multiple of the wavelength. The given dark fringe for \(m=0\) is at \(\theta=15^{\circ}\). We need to find the angle \(\theta\) for \(m=1\).
02

Identifying the Formula

The condition for dark fringes in a double-slit experiment is given by: \( d \sin(\theta) = (m + 0.5) \lambda \), where \(d\) is the slit separation, \(\lambda\) is the wavelength, and \(m\) is the order of the fringe.
03

Applying for m=0

For \(m=0\), the angle \(\theta = 15^{\circ}\). The condition becomes: \( d \sin(15^{\circ}) = 0.5 \lambda \). Since we don't have \(d\) and \(\lambda\), we'll use the formula to express these parameters in terms of known variables.
04

Finding the Angle for m=1

For \(m=1\), substitute the values into the formula: \( d \sin(\theta) = (1 + 0.5) \lambda = 1.5 \lambda \). By comparing the equations \( d \sin(15^{\circ}) = 0.5 \lambda \) and \( d \sin(\theta) = 1.5 \lambda \), the relationship is: \( \sin(\theta) = 3 \sin(15^{\circ}) \). Since this result is unachievable (sin of an angle cannot be greater than 1), hypothesize based on typical result contexts.
05

Conclusion and Angle Adjustment

Usually when computing for angles where \( \sin(\theta) > 1 \), it implies reaching a region with no solution or invalid physical scenario under given experimental bars. Hence, the formula must have constraints. Given expectations in typical regions, it's understood potentially an incorrect angle occurred or experiment-barrier in angle; preferably re-check method or hypothesis utilizing \(\sin^{-1}(\text{expected region})\) for confirmation, in typical model, contrasting circle behaviors.

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Key Concepts

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

Interference Patterns
In Young's double-slit experiment, interference patterns are created when light waves from two separate slits overlap and combine. This combination leads to a series of bright and dark bands on a screen, a phenomenon known as interference. These patterns occur due to constructive and destructive interference. Constructive interference happens when the waves align in such a way that their amplitudes add together, producing bright fringes.

On the other hand, destructive interference occurs when the peaks of one wave match the troughs of another, canceling each other out and resulting in dark fringes. The regular set of light and dark bands depends on the wavelength of light and the distance between the slits. The position and spacing of the fringes provide valuable information about the wave character of light. Thus, these interference patterns are a visual representation of the wave nature of light.
  • Constructive Interference: Results in bright fringes
  • Destructive Interference: Causes dark fringes
  • Information Source: The pattern shows the wave nature of light
Dark Fringes
Dark fringes are specific points in the interference pattern where destructive interference occurs. In the context of Young's double-slit experiment, they appear as dark bands on the projection screen. These occur where the path difference between waves from the two slits is equal to a half-integer multiple of the wavelength (e.g., \(m + 0.5\)\( \lambda \)).

The concept of fringes can be quite fascinating, as it demonstrates the inherent wave properties of light, being able to cancel when out of phase. Each dark fringe corresponds with an interference order, denoted with integer values like \(m\). For instance, \(m=0\) indicates the first order dark fringe (also referred to as a primary dark fringe).

When calculating these positions, sine and cosine functions are often used due to the angular nature of the interference pattern. This is a critical concept in physics, enhancing the understanding of wave behaviors.
  • Occurs due to: Destructive interference
  • Calculation: Based on path difference equal to \(m + 0.5\)\( \lambda \)
  • Significance: Demonstrates wave properties of light
Angle of Diffraction
The angle of diffraction in Young's double-slit experiment refers to the specific angle at which the light is bent as it passes through the slits and creates interference patterns. This angle helps determine where the dark and bright fringes appear on the screen.

In mathematical terms, the angle of diffraction is crucial for locating dark fringes using the formula \(d \sin(\theta) = (m + 0.5) \lambda\). Here, \(d\) represents the distance between the two slits, \(\theta\) is the angle of diffraction, \(m\) is the fringe order, and \(\lambda\) is the wavelength of the light.

Interestingly, in real-world experiments, sometimes the sine of the angle can approach or exceed 1, which is mathematically unsolvable as sine values must be between -1 and 1. This indicates that practical constraints, such as the distance between slits, need reconsideration if results fall out of the range. Therefore, understanding the angle of diffraction deepens our comprehension of physical limits in experimental setups.
  • Purpose: Determines fringe positions
  • Relation: \(d \sin(\theta) = (m + 0.5) \lambda\)
  • Limitations: Sine of angle must be between -1 and 1

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Most popular questions from this chapter

The distance between adjacent slits of a certain diffraction grating is \(1.250 \times 10^{-5} \mathrm{m}\) . The grating is illuminated by monochromatic light with a wavelength of \(656.0 \mathrm{nm},\) and is then heated so that its temperature increases by 100.0 \(\mathrm{C}^{\circ} .\) Determine the change in the angle of the seventh order principal maximum that occurs as a result of thermal expansion of the grating. The cofficient of linear expansion for the diffaction grating is \(1.30 \times 10^{-4}\left(\mathrm{C}^{9}\right)^{-1} .\) Be sure to include the proper algebraic sign with your answer: \(+\) if the angle increases, - if the angle decreases.

Two concentric circles of light emit light whose wavelength is 555 nm. The larger circle has a radius of 4.0 cm, and the smaller circle has a radius of 1.0 cm. When taking a picture of these lighted circles, a camera admits light through an aperture whose diameter is 12.5 mm. What is the maximum distance at which the camera can (a) distinguish one circle from the other and (b) reveal that the inner circle is a circle of light rather than a solid disk of light?

A flat screen is located 0.60 m away from a single slit. Light with a wavelength of 510 nm (in vacuum) shines through the slit and produces a diffraction pattern. The width of the central bright fringe on the screen is 0.050 m. What is the width of the slit?

In a single-slit diffraction pattern, the central fringe is 450 times as wide as the slit. The screen is 18 000 times farther from the slit than the slit is wide. What is the ratio \(\lambda / W,\) where \(\lambda\) is the wavelength of the light shining through the slit and \(W\) is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that \(\sin \theta \approx \tan \theta\)

A flat observation screen is placed at a distance of 4.5 m from a pair of slits. The separation on the screen between the central bright fringe and the first-order bright fringe is 0.037 m. The light illuminating the slits has a wavelength of 490 nm. Determine the slit separation
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