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Chapter 36: Problem 1

The distance between the first and fifth minima of a single-slit diffraction pattern is \(0.35 \mathrm{~mm}\) with the screen \(40 \mathrm{~cm}\) away from the slit, when light of wavelength \(550 \mathrm{nm}\) is used. (a) Find the slit width. (b) Calculate the angle \(\theta\) of the first diffraction minimum.

Short Answer

Expert verified
The slit width is 25.14 碌m, and the angle \(\theta\) is approximately \(1.255^\circ\).

Step by step solution

01

Understand the Diffraction Formula

For single-slit diffraction, the position of the m-th minimum on a screen is given by the formula \(y_m = \frac{m \lambda L}{a}\), where \(m\) is the order of the minimum, \(\lambda\) is the wavelength of light, \(L\) is the distance to the screen, and \(a\) is the slit width. We are given the distance between the first (\(m=1\)) and fifth (\(m=5\)) minima, which is \(y_5 - y_1 = 0.35 \ \text{mm}\).
02

Set Up the Equation for Slit Width

Use the equation for the position of the minima: \(y_5 - y_1 = \frac{5 \cdot \lambda \cdot L}{a} - \frac{1 \cdot \lambda \cdot L}{a} = \frac{4 \cdot \lambda \cdot L}{a}\). We can use this equation to solve for the slit width \(a\).
03

Substitute Given Values and Solve for Slit Width

Substitute the given values into the equation: \(0.35 \times 10^{-3} = \frac{4 \cdot 550 \times 10^{-9} \cdot 0.4}{a}\). Simplifying gives: \(a = \frac{4 \cdot 550 \times 10^{-9} \cdot 0.4}{0.35 \times 10^{-3}} = 2.514\times 10^{-5} \ \text{m}\; or \; 25.14 \ \text{碌m}\).
04

Calculate the Angle for the First Diffraction Minimum

The angle \(\theta\) for the first minimum\((m=1)\) is given by \(a \sin \theta = m \lambda\), which simplifies to \(\sin \theta = \frac{\lambda}{a}\). Substitute the known values: \(\sin \theta = \frac{550 \times 10^{-9}}{2.514\times 10^{-5}}\). Calculate \(\theta\) using the inverse sine function.
05

Solve for \(\theta\)

Using a calculator, find \(\theta = \sin^{-1}\left(\frac{550 \times 10^{-9}}{2.514\times 10^{-5}}\right)\). The resulting angle is approximately \(1.255^\circ\).

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Key Concepts

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

Diffraction Minima
In single-slit diffraction, understanding the formation of minima is crucial. When light waves pass through a narrow slit, they spread out, creating a pattern of bright and dark regions on a screen. These dark regions are known as diffraction minima. They occur due to destructive interference where wave amplitudes cancel each other out.
  • The position of these minima depends on the order number \(m\), the wavelength \(\lambda\), the width of the slit \(a\), and the distance \(L\) to the screen.
  • The order of the minimum \(m\) refers to its sequence in the pattern, starting with \(m=1\) for the first minimum.
  • The formula to find the position \(y_m\) of the m-th minimum is: \(y_m = \frac{m \lambda L}{a}\).
This relationship helps us calculate specific properties of the diffraction pattern, such as slit width if the distance between minima is known.
Wavelength of Light
The wavelength of light is a key factor in diffraction phenomena. It represents the distance between consecutive peaks of a wave of light, usually measured in nanometers (nm).
  • In the given problem, the wavelength \(\lambda\) is \(550\ nm\), which is in the visible spectrum of light, giving it a distinctive color.
  • It influences how light interacts with the slit, affecting the position and width of the diffraction pattern on the screen.
  • The longer the wavelength, the more pronounced the diffraction effect, because longer wavelengths spread out more as they diffract through a slit.
Understanding wavelengths is crucial for predicting and analyzing diffraction behaviors in optical systems.
Slit Width
The slit width \(a\) in diffraction experiments plays a vital role in the resulting pattern's characteristics. It refers to the opening through which light passes before spreading out.
  • In the exercise, the slit width is calculated using the formula derived from the diffraction minima definition: \(y_5 - y_1 = \frac{4 \lambda L}{a}\).
  • A narrower slit leads to a broader and more spaced-out diffraction pattern, highlighting more visible minima and maxima.
  • Conversely, a wider slit will result in a tighter pattern with less distinct features.
Determining the slit width is essential for many applications in optics, such as designing optical instruments and understanding wave behavior.
Angle of Diffraction
The angle of diffraction \(\theta\) is a measure of how much the light wave deviates from its original path after passing through the slit. This angle is crucial for characterizing the diffraction pattern.
  • The angle is calculated using the relationship \(a \sin \theta = m \lambda\), where \(m\) is the order of the minimum.
  • Knowing this angle helps in predicting where the dark and bright regions will appear on the screen.
  • In practical terms, with the given problem, this angle allows us to determine the exact placement of diffraction minima.
Calculating \(\theta\) gives us insight into the intensity distribution across the diffraction pattern, serving important roles in optical engineering and wave analysis.

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Most popular questions from this chapter

An \(x\) -ray beam of wavelengths from 95.0 to \(140 \mathrm{pm}\) is incident at \(\theta=45.0^{\circ}\) to a family of reflecting planes with spacing \(d=275 \mathrm{pm}\). What are the (a) longest wavelength \(\lambda\) and (b) associated order number \(m\) and the (c) shortest \(\lambda\) and (d) associated \(m\) of the intensity maxima in the diffraction of the beam?

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{nm}\). The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) What is the ratio of the beam intensity at the target to that at the end of the wire? (The laser is fired from space, so neglect any atmospheric absorption.)

In a double-slit experiment, the slit separation \(d\) is 2.00 times the slit width \(w\). How many bright interference fringes are in the central diffraction envelope?

If first-order reflection occurs in a crystal at Bragg angle \(3.4^{\circ}\) at what Bragg angle does second-order reflection occur from the same family of reflecting planes?

In conventional television, signals are broadcast from towers to home receivers. Even when a receiver is not in direct view of a tower because of a hill or building, it can still intercept a signal if the signal diffracts enough around the obstacle, into the obstacle's "shadow region." Previously, television signals had a wavelength of about \(50 \mathrm{~cm}\), but digital television signals that are transmitted from towers have a wavelength of about \(10 \mathrm{~mm}\). (a) Did this change in wavelength increase or decrease the diffraction of the signals into the shadow regions of obstacles? Assume that a signal passes through an opening of \(5.0 \mathrm{~m}\) width between two adjacent buildings. What is the angular spread of the central diffraction maximum (out to the first minima) for wavelengths of (b) \(50 \mathrm{~cm}\) and (c) \(10 \mathrm{~mm} ?\)
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