Problem 54 - University Physics with Modern Physics

Question

A slit 0.360 mm wide is illuminated by parallel rays of light that have a wavelength of 540 nm. The diffraction pattern is observed on a screen that is 1.20 m from the slit. The intensity at the center of the central maximum \((\theta = 0^\circ)\) is \(I_0\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_0\)/2?

Step-by-Step Solution

Answer

(a) The first minimum is 1.80 mm from the center; (b) the intensity is half at 0.675 mm.

Step 1: Convert Units and Define Variables

First, convert the width of the slit and the wavelength into meters: \(a = 0.360 \, \text{mm} = 0.360 \times 10^{-3} \, \text{m}\) and \(\lambda = 540 \, \text{nm} = 540 \times 10^{-9} \, \text{m}\). The distance from the slit to the screen is \(L = 1.20 \, \text{m}\).

Step 2: Apply the Formula for the First Minimum

The formula for the position of the first diffraction minimum is \(a \sin \theta = m \lambda\), where \(m = 1\) for the first minimum. Therefore, \(\sin \theta = \frac{\lambda}{a}\). Substitute: \(\sin \theta = \frac{540 \times 10^{-9}}{0.360 \times 10^{-3}} = 1.5 \times 10^{-3}\).

Step 3: Calculate the Angle for the First Minimum

Since \(\theta\) is small, we can approximate \(\sin \theta \approx \theta\) in radians. So, \(\theta = 1.5 \times 10^{-3} \, \text{radians}\).

Step 4: Find the Distance on the Screen for the First Minimum

Use the small angle approximation formula for distance from the center: \(y = L \tan \theta \approx L \theta\). Therefore, \(y = 1.20 \times 1.5 \times 10^{-3} = 1.80 \times 10^{-3} \, \text{m}\) or \(1.80 \, \text{mm}\).

Step 5: Formula for Intensity (Half of Central Maximum)

The intensity at a point where it is half of the central maximum is given by \(I = I_0 \sin^2 \beta\). To find where \( I = \frac{I_0}{2}\), we use \(\sin^2 \beta = \frac{1}{2}\) which means \(\beta = \frac{3\pi}{8}\) (first solution such that \(\sin \, \beta = \frac{1}{\sqrt{2}}\)).

Step 6: Relate Beta to Slit Parameters

The term \(\beta = \frac{\pi a \sin \theta}{\lambda}\) relates \(\beta\) to the slit parameters. Therefore, \(\frac{\pi \cdot 0.360 \times 10^{-3} \sin \theta}{540 \times 10^{-9}} = \frac{3\pi}{8}\) simplifies to \(\sin \theta \approx \frac{3 \cdot 540 \times 10^{-9}}{8 \cdot 0.360 \times 10^{-3}} \approx 5.625 \times 10^{-4}\).

Step 7: Calculate the Distance Where Intensity is I_0/2

For small angles, \(\theta = 5.625 \times 10^{-4}\) radians. Thus, \(y = L \cdot \theta = 1.20 \cdot 5.625 \times 10^{-4} = 6.75 \times 10^{-4} \, \text{m}\) or \(0.675 \, \text{mm}\).

Key Concepts

Single Slit Diffraction

Diffraction is a fascinating phenomenon that occurs when waves encounter an obstacle or aperture. In optics, this is typically seen when light beams pass through a small slit. One of the fundamental ways this happens is known as single slit diffraction.

• The slit causes the light waves to spread out and interfere with each other.
• This interference leads to a pattern of bright and dark fringes on a screen, known as a diffraction pattern.

In single slit diffraction, the central bright fringe (known as the central maximum) is the widest and brightest. As you move away from the center, the intensity of the light diminishes, creating alternating dark and light regions.
These dark regions, or minima, occur when the light waves are out of phase, causing destructive interference. The formula for locating these minima uses the relationship: \[ a \sin \theta = m \lambda \]where:

• \(a\) is the width of the slit,
• \(\theta\) is the angle relative to the central maximum,
• \(m\) is the order of the minimum, starting from \(m=1\) for the first minimum,
• \(\lambda\) is the wavelength of the light used.

Intensity Distribution

In single slit diffraction, the intensity of the light on the screen doesn't remain constant. It varies due to the interference of the light waves, forming a phenomenon known as intensity distribution.
The central maximum is the brightest part of the diffraction pattern, where the intensity is at its peak. As we move away from this point, the intensity dips and climbs periodically.
To understand at what points this intensity diminishes significantly, we use the formula:\[ I(\theta) = I_0 \left(\frac{\sin \beta}{\beta}\right)^2 \]For practical purposes, when we want to find the point where the intensity is half of the central maximum value \(I_0\), we'd use the specific condition:

• \(\beta = \frac{3\pi}{8}\) is derived from solving \(\sin^2\beta = \frac{1}{2}\).

This tells us the angle \(\theta\) at which the intensity falls to half its maximum value, allowing us to predict the location of the half-maximum intensity point on the screen using the angle \(\theta\).
This knowledge of intensity distribution helps in understanding places of constructive and destructive interference in the light pattern.

Wavelength

Wavelength is a pivotal concept in understanding diffraction patterns. It refers to the distance between consecutive peaks or troughs in a wave. In the context of light, it determines the color in the visible spectrum, with common units including nanometers (nm).
When light passes through a narrow slit, its wavelength determines the nature of the diffraction pattern.

• A longer wavelength (like red light) will produce a wider diffraction pattern.
• Conversely, a shorter wavelength (like blue light) will produce a narrower pattern.

The relationship between the slit width \(a\), the order of the minimum \(m\), and the wavelength \(\lambda\) helps locate the minima positions in the diffraction pattern: \[ a \sin \theta = m \lambda \]This equation is critical because it allows us to determine the physical spacing of the dark and bright fringes based on the wavelength of the light used. In practical exercises, such as the one provided, the wavelength is crucial for calculating angles and distances necessary to understand and predict the diffraction pattern behavior.
Thus, wavelength not only influences the diffraction process but is also instrumental in analyzing and predicting the results of diffraction phenomena.