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Chapter 36: Problem 7

Light of wavelength \(633 \mathrm{nm}\) is incident on a narrow slit. The angle between the first diffraction minimum on one side of the central maximum and the first minimum on the other side is \(1.20^{\circ} .\) What is the width of the slit?

Short Answer

Expert verified
The slit width is approximately 60.5 micrometers.

Step by step solution

01

Identify the formula for the diffraction minimum

The angle for the first diffraction minimum from a single slit is given by the formula \( a \sin\theta = m\lambda \), where \(a\) is the slit width, \(\theta\) is the angle for the minimum, \(m\) is the order of the minimum, and \(\lambda\) is the wavelength of the light. For the first minimum, \(m = 1\).
02

Convert the angle to radians

The angle between the first minima on either side of the central maximum is given in degrees. Since the angle for one side (either positive or negative) is needed, we use \(\theta = \frac{1.20^{\circ}}{2} = 0.60^{\circ}\). Convert this to radians using \(1 \, \text{degree} = \frac{\pi}{180} \, \text{radians}\), so \(\theta = 0.60^{\circ} \times \frac{\pi}{180} = 0.01047 \, \text{radians}\).
03

Rearrange the formula to solve for slit width

Rearrange \( a \sin\theta = m\lambda \) to solve for \(a\): \( a = \frac{m\lambda}{\sin\theta} \). For the first minimum, \(m = 1\). Thus, \( a = \frac{\lambda}{\sin\theta} \).
04

Substitute the known values into the formula

Substitute \(\lambda = 633 \, \text{nm} = 633 \times 10^{-9} \, \text{m}\) and \(\theta = 0.01047\) radians into the formula: \( a = \frac{633 \times 10^{-9}}{\sin(0.01047)} \).
05

Calculate the slit width

Calculate \(\sin(0.01047)\) which is approximately \(0.01047\) as the angle is very small. Therefore, \( a = \frac{633 \times 10^{-9}}{0.01047}\). Calculate this to find \(a \approx 6.05 \times 10^{-5} \, \text{m} \) or \(60.5 \, \mu\text{m}\).

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Key Concepts

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

Diffraction Pattern
When light passes through a narrow slit, it doesn't simply continue in a straight line. Instead, it spreads out, creating regions of light and dark known as a diffraction pattern.
This fascinating phenomenon occurs due to the wave nature of light. Understanding the diffraction pattern:
  • The central band is the brightest and widest, known as the central maximum.
  • On either side of the central maximum, bands of decreasing intensity, known as maxima and minima, appear.
  • Minima are points where the light waves cancel each other due to interference, resulting in dark regions.
The angle between these minima helps in determining crucial parameters like slit width in an experimental setup.
Wavelength
Wavelength is a fundamental property of waves, including light waves. It refers to the distance between consecutive crests or troughs in a wave.Why Wavelength Matters:
  • In the context of diffraction, wavelength (\( \lambda \)) determines how much the light spreads out when it encounters a slit.
  • The color of light is associated with its wavelength. For instance, the wavelength mentioned in this exercise, 633 nm, corresponds to red light in the spectrum.
Knowing the wavelength is crucial in calculations related to wave behavior, such as finding the slit width using the diffraction pattern.
Angular Measurement
In the study of diffraction, we often measure angles to identify points of maxima and minima. Angular measurement helps in analyzing how waves interact over distances.Understanding the Angle in Diffraction:
  • The angle \( \theta \)) is often measured in degrees but must be converted into radians for most physics equations.
  • In our exercise, the angle between the first diffraction minima on either side of the central maximum was provided as 1.20掳. To use it in calculations, we specifically handled the angle for one side, 0.60掳, and converted it to radians (0.01047 radians).
  • This precise measurement is vital for calculating factors such as slit width using appropriate formulas.
Slit Width Calculation
Calculating the width of a slit in a diffraction experiment allows us to interpret how light behaves when passing through narrow openings.Step-by-Step Calculation:
  • The formula used is: \( a \sin\theta = m\lambda \), where \( a \) is the slit width, \( m \) is the order of the minimum (1 for the first minimum), and \( \lambda \) is the wavelength.
  • To find the slit width \( a \), rearrange the formula to \( a = \frac{\lambda}{\sin\theta} \).
  • Substitute known values: \( \lambda = 633 \times 10^{-9} \, \text{m} \), \( \theta = 0.01047 \, \text{radians} \).
  • For small angles, \( \sin\theta \approx \theta \), so \( a = \frac{633 \times 10^{-9}}{0.01047} \).
  • Calculate the value to find \( a \approx 6.05 \times 10^{-5} \, \text{m} \) or \( 60.5 \mu\text{m} \). This is the width of the slit.
This method, along with understanding the underlying concepts, provides a clear path to experimentally determining slit dimensions in optics.

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Most popular questions from this chapter

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) \(10 \mathrm{~mm} ?\)

(A) How far from grains of red sand must you be to position yourself just at the limit of resolving the grains if your pupil diameter is \(1.5 \mathrm{~mm}\), the grains are spherical with radius \(50 \mu \mathrm{m}\), and the light from the grains has wavelength \(650 \mathrm{nm} ?\) (b) If the grains were blue and the light from them had wavelength \(400 \mathrm{nm}\), would the answer to (a) be larger or smaller?

With a particular grating the sodium doublet \((589.00 \mathrm{nm}\) and \(589.59 \mathrm{nm}\) ) is viewed in the third order at \(10^{\circ}\) to the normal and is barely resolved. Find (a) the grating spacing and (b) the total width of the rulings.

In the single-slit diffraction experiment of Fig. \(36-4,\) let the wavelength of the light be \(500 \mathrm{nm}\), the slit width be \(6.00 \mu \mathrm{m}\), and the viewing screen be at distance \(D=3.00 \mathrm{~m}\). Let a \(y\) axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let \(I_{P}\) represent the intensity of the diffracted light at point \(P\) at \(y=15.0 \mathrm{~cm} .\) (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where point \(P\) is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

The wall of a large room is covered with acoustic tile in which small holes are drilled \(5.0 \mathrm{~mm}\) from center to center. How far can a person be from such a tile and still distinguish the individual holes, assuming ideal conditions, the pupil diameter of the observer's eye to be \(4.0 \mathrm{~mm},\) and the wavelength of the room light to be \(550 \mathrm{nm} ?\)
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