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Chapter 36: Problem 15

A slit \(0.240 \mathrm{~mm}\) wide is illuminated by parallel light rays of wavelength \(540 \mathrm{nm} .\) The diffraction pattern is observed on a screen that is \(3.00 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

Short Answer

Expert verified
(a) 6.75 mm; (b) 5.74 x 10^-6 W/m^2.

Step by step solution

01

Determine First Minimum Condition

In a single-slit diffraction pattern, the condition for the first minimum is given by the equation \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle relative to the central maximum, \( m \) is the order of the minimum (for the first minimum, \( m = 1 \)), and \( \lambda \) is the wavelength of the light. Plugging in the values, we have:\[ 0.240 \times 10^{-3} \sin \theta = 1 \times 540 \times 10^{-9} \].Solve for \( \sin \theta \).
02

Calculate the Angle of the First Minimum

From the equation in Step 1, solve for \( \sin \theta \):\[ \sin \theta = \frac{540 \times 10^{-9}}{0.240 \times 10^{-3}} = 2.25 \times 10^{-3} \].Since \( \sin \theta \) is small, we can approximate \( \theta \approx \sin \theta \). So, \( \theta \approx 2.25 \times 10^{-3} \text{ rad} \).
03

Determine Distance to First Minimum on Screen

The linear distance \( y \) to the first minimum on the screen is calculated using the small angle approximation \( y = L \tan \theta \approx L \sin \theta \), where \( L = 3.00 \text{ m} \). Thus:\[ y = 3.00 \times 2.25 \times 10^{-3} = 6.75 \times 10^{-3} \text{ m} \].
04

Find Midway Position

To find the midway position between the center of the central maximum and the first minimum, take half of the distance calculated in Step 3:\[ y_{midway} = \frac{6.75 \times 10^{-3}}{2} = 3.375 \times 10^{-3} \text{ m} \].
05

Calculate Intensity at Midway

The intensity \( I \) at a point on the screen in a single-slit diffraction pattern is given by:\[ I = I_0 \left( \frac{\sin \beta}{\beta} \right)^2 \]where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( I_0 \) is the intensity at the center. For the midway point, \( \theta_{midway} = \frac{\theta}{2} = 1.125 \times 10^{-3} \text{ rad} \). Calculate \( \beta \) and substitute into the intensity formula:\[ \beta = \frac{\pi \times 0.240 \times 10^{-3} \times 1.125 \times 10^{-3}}{540 \times 10^{-9}} \approx 0.48 \].\( I = 6.00 \times 10^{-6} \left( \frac{\sin(0.48)}{0.48} \right)^2 \).Calculate this to find \( I \approx 5.74 \times 10^{-6} \text{ W/m}^2 \).

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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 single slit, it spreads out and creates a pattern of bright and dark areas on a screen. This is the diffraction pattern. It's like when you squeeze water through a narrow opening; the water spreads out on the other side.

The pattern consists of a series of light and dark bands. The brightest spot, known as the central maximum, appears directly opposite the slit. As you move away from the central maximum, the light intensity decreases, leading to darker areas known as minima. These alternating bright and dark regions form because the light waves interfere with each other鈥攕ometimes they add up, creating bright spots, and sometimes they cancel out, forming dark spots.
Intensity Distribution
In a diffraction pattern, not all light is distributed evenly. The intensity, or brightness, of the light changes as you move across the pattern.

The intensity is highest at the center, known as the central maximum, and then it decreases as you move further away. This decrease happens because the waves are not perfectly in sync everywhere; they interfere with each other, lessening their combined brightness. The varying intensity creates the light and dark bands in the pattern.

This distribution is mathematically described by a formula involving the sin function: - At the center, the intensity is maximum because all light waves add up constructively. - Towards the edges, destructive interference reduces the light intensity, producing darker regions.
Central Maximum
The central maximum is the brightest part of a diffraction pattern. It's located directly opposite to where the light enters the slit. This occurs because, at this point, all the light waves traveling through the slit combine their peaks and troughs in perfect harmony.

Unlike other parts of the pattern, the central maximum is usually much brighter and wider. It's also the reference for measuring other parts of the pattern, such as the first and other minima. In the original problem, the central maximum's intensity is the starting point for calculating the intensity at other points in the pattern. Knowing it helps us understand how the light behaves as it travels through the slit and onto the screen.
Light Wavelength
The wavelength of light refers to the distance between two peaks of a light wave. It's a crucial factor in determining the characteristics of the diffraction pattern.

In our exercise, the light's wavelength is given as 540 nm (nanometers). This value influences the spacing between the diffraction minima and maxima. The longer the wavelength, the more spread out the diffraction pattern becomes.

The calculation for the diffraction pattern, such as finding the position of minimum intensity on the screen, directly involves the wavelength. When the light passes through the slit, the path difference caused depends on the wavelength, affecting how the waves interfere with each other.

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Most popular questions from this chapter

Monochromatic x rays are incident on a crystal for which the spacing of the atomic planes is 0.440 nm. The first-order maximum in the Bragg reflection occurs when the incident and reflected x rays make an angle of 39.4\(^\circ\) with the crystal planes. What is the wavelength of the x rays?

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm\)90.0\(^\circ\), so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta\) = 45.0\(^\circ\) to the intensity at \(\theta\) = 0?

A laser beam of wavelength \(\lambda\) = 632.8 nm shines at normal incidence on the reflective side of a compact disc. (a) The tracks of tiny pits in which information is coded onto the CD are 1.60 \(\mu\)m apart. For what angles of reflection (measured from the normal) will the intensity of light be maximum? (b) On a DVD, the tracks are only 0.740 \(\mu\)m apart. Repeat the calculation of part (a) for the DVD.

Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many \(totally\) dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem \(without\) calculating all the angles! (\(Hint\): What is the largest that sin \theta can be? What does this tell you is the largest that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

Monochromatic light with wavelength 490 nm passes through a circular aperture, and a diffraction pattern is observed on a screen that is 1.20 m from the aperture. If the distance on the screen between the first and second dark rings is 1.65 mm, what is the diameter of the aperture?
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