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Chapter 36: Problem 51

The distance between the first and fifth minima of a single-slit diffraction pattern is \(0.50 \mathrm{~mm}\) with the screen \(40 \mathrm{~cm}\) away from the slit, when light of wavelength \(420 \mathrm{~nm}\) is used. (a) Find the slit width. (b) Calculate the angle \(\theta\) of the first diffraction minimum.

Short Answer

Expert verified
Slit width \(a\) is 0.1344 mm. Angle \(\theta_1\) is approximately 0.179 degrees.

Step by step solution

01

Understand the Problem

Identify what is given and what we need to find. We are given the distance between the first and fifth minima, the distance from the screen to the slit, and the wavelength of the light. We need to find the slit width and the angle at the first minimum.
02

Apply the Single-Slit Diffraction Formula

For a single-slit diffraction pattern, the position of the minima is given by the formula \( a \sin\theta_m = m \lambda \), where \( a \) is the slit width, \( \theta_m \) is the angle to the \( m^{th} \) minimum, \( m \) is the minimum order, and \( \lambda \) is the wavelength.
03

Setup the Equation for the Distance between Minima

The distance between the first and fifth minima (4th-barring) on the screen is given by \( x = L(\tan\theta_5 - \tan\theta_1) \). For small angles (which is common in diffraction), \( \tan\theta \approx \sin\theta \approx \theta \). Thus, \( x \approx L(\theta_5 - \theta_1) \).
04

Relate \( \theta \) Using the Diffraction Angle Formula

For the minima, \( \theta_m = \frac{m\lambda}{a} \). So, the distance is expressed as: \[ x = L\left(\frac{5\lambda}{a} - \frac{\lambda}{a}\right) = L\frac{4\lambda}{a} \]
05

Solve for Slit Width \( a \)

Rearrange the formula to solve for \( a \): \[ a = \frac{4\lambda L}{x} \] Substitute the known values: \( \lambda = 420 \times 10^{-9} \ m \), \( L = 0.4 \ m \), \( x = 0.50 \times 10^{-3} \ m \).
06

Calculate Slit Width

Substitute the values into the equation:\[ a = \frac{4 \times 420 \times 10^{-9} \times 0.4}{0.50 \times 10^{-3}} = 1.344 \times 10^{-4} \ m = 0.1344 \ mm \]
07

Determine the Angle \( \theta \) for First Minimum

Using the formula \( \theta_1 = \frac{\lambda}{a} \), substitute \( \lambda = 420 \times 10^{-9} \ m \) and the calculated slit width \( a = 1.344 \times 10^{-4} \ m \).
08

Perform Calculation for \( \theta_1 \)

Calculate the angle:\[ \theta_1 = \frac{420 \times 10^{-9}}{1.344 \times 10^{-4}} = 3.125 \times 10^{-3} \ radians \] Convert to degrees if needed: approximately \( 0.179 \) degrees.

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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 spreads out and creates a pattern of bright and dark fringes on a screen. This phenomenon is known as a diffraction pattern. It is caused by constructive and destructive interference, where waves coming from different parts of the slit interfere with each other. The central bright region is the most intense, with alternating dark and bright fringes on either side. The dark fringes correspond to diffraction minima, where destructive interference occurs. Understanding the arrangement of these fringes is key in solving problems involving single-slit diffraction, as it provides a visual representation of how light behaves when encountering a barrier.
Slit Width Calculation
Determining the width of the slit in a diffraction experiment involves using the positions of the minima. For a single-slit diffraction pattern, the formula is given by:\[ a \sin\theta_m = m \lambda \]where:
  • \(a\) is the slit width.
  • \(\theta_m\) is the angle at the \(m\)-th minimum.
  • \(m\) is the order of the minimum.
  • \(\lambda\) is the wavelength of light.
In practical applications, we might calculate the slit width using the distance between the minima, which can be measured on a screen placed behind the slit.The formula we use to relate this is:\[ x = L \left( \frac{5\lambda}{a} - \frac{\lambda}{a} \right) \]This is rearranged to solve for the slit width:\[ a = \frac{4\lambda L}{x} \]By substituting known values for the wavelength, distance to the screen, and distance between the minima, we can accurately calculate the slit width.
Angle of Diffraction Minimum
The angle at which these minima occur is crucial in understanding diffraction patterns. For any minimum, this angle \(\theta\) can be calculated using the formula:\[ \theta_m = \frac{m\lambda}{a} \]For the first minimum, which is often of primary interest, the angle \(\theta_1\) is calculated as:\[ \theta_1 = \frac{\lambda}{a} \]This formula shows the direct relationship between the wavelength of light, the slit width, and the resulting angle of the dark fringe. It's important to remember that for small angles often encountered in diffraction experiments, \(\sin\theta \approx \theta\), simplifying calculations.When working through these problems, it鈥檚 common to express angles in radians, providing ease of calculation, though conversion into degrees may be needed for practical interpretations or comparison with empirical data.

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Most popular questions from this chapter

A single slit is illuminated by light of wavelengths \(\lambda_{a}\) and \(\lambda_{\text {on }}\) chosen so that the first diffraction minimum of the \(\lambda_{2}\) component coincides with the second minimum of the \(\lambda_{b}\) component. (a) If \(\lambda_{b}=420 \mathrm{~nm}\), what is \(\lambda_{\alpha}\) ? For what order number \(m_{b}\) (if any) does a minimum of the \(\lambda_{b}\) component coincide with the minimum of the \(\lambda_{a}\) component in the order number (b) \(m_{a}=2\) and (c) \(m_{a}=3\) ?

Find the separation of two points on the Moon's surface that can just be resolved by the \(200 \mathrm{in} .(=5.1 \mathrm{~m})\) telescope at Mount Palomar, assuming that this separation is determined by diffraction effects. The distance from Earth to the Moon is \(3.8 \times 10^{5} \mathrm{~km}\). Assume a wavelength of \(550 \mathrm{~nm}\) for the light.

A grating has 600 rulings/mm and is \(6.0 \mathrm{~mm}\) wide. (a) What is the smallest wavelength interval it can resolve in the third order at \(\lambda=500 \mathrm{~nm}\) ? (b) How many higher orders of maxima can be seen?

Light at wavelength \(589 \mathrm{~nm}\) from a sodium lamp is incident perpendicularly on a grating with 40000 rulings over width \(96 \mathrm{~mm}\). What are the first-order (a) dispersion \(D\) and (b) resolving power \(R\), the second-order (c) \(D\) and (d) \(R\), and the third-order (e) \(D\) and (f) \(R\) ?

A diffraction grating \(24.0 \mathrm{~mm}\) wide has 5000 rulings. Light of wavelength \(589 \mathrm{~nm}\) is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of \(\theta\) at which maxima appear on a distant viewing screen?
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