Question

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?

Short answer

(a) Slit width is 580 nm. (b) Intensity ratio is 0.405.

Step-by-step solution

Understanding the Problem

We are asked to calculate two main things. First, the width of a slit given that the first diffraction minima are at 90.0° (which means the central maximum spans the entire screen). Second, we need to find the intensity ratio at two specific angles for that slit width.

Using Formula for First Minima in Single-Slit Diffraction

The condition for the first diffraction minima in a single-slit diffraction pattern is given by \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle of diffraction (90° in this case), \( m \) is the order of the minima (1 for the first minima), and \( \lambda \) is the wavelength of light (580 nm).

Calculating the Slit Width

Substitute the given values into the formula: \( a \sin 90.0° = 1 \times 580 \times 10^{-9} \), where \( \sin 90.0° = 1 \). Therefore, \( a = 580 \times 10^{-9} \text{ m} = 580 \, \text{nm}\). The width of the slit is 580 nm.

Apply Intensity Formula for Single-Slit Diffraction

The relative intensity for a single slit is given by \( I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \), where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( I_0 \) is the intensity at \( \theta = 0 \).

Calculate Intensity at \( \theta = 45.0° \)

Substitute \( \theta = 45.0° \) into \( \beta \):\[ \beta = \frac{\pi \times 580 \times 10^{-9} \times \sin 45.0°}{580 \times 10^{-9}} = \frac{\pi}{\sqrt{2}} \]Next, calculate \( \left(\frac{\sin(\beta)}{\beta}\right)^2 \) at \( \beta = \frac{\pi}{\sqrt{2}} \).

Calculate Intensity Ratio

Calculate \( \left(\frac{\sin(\pi/\sqrt{2})}{\pi/\sqrt{2}}\right)^2 \) to find the ratio. The term simplifies to approximately 0.405. Therefore, the intensity at \( \theta = 45.0° \) is 40.5% of the intensity at \( \theta = 0\). The ratio is 0.405.

Key concepts

Single-slit diffraction

When light passes through a narrow slit, it doesn't just continue straight. Instead, it spreads out in a phenomenon known as diffraction. This is more noticeable when the slit is comparable in size to the wavelength of light. One of the simplest setups to observe this is in single-slit diffraction. Here, a single narrow slit is used to diffract light onto a screen, creating a pattern of light and dark areas due to the interference of light waves.

Single-slit diffraction is a key example of Fraunhofer diffraction. This type of diffraction occurs when both the light source and the observation screen are far enough from the slit. In such cases, the light waves can be treated as parallel rays. The central bright area on the screen is flanked by a series of dark and bright fringes.

These patterns arise because parts of the light waves overlap and interfere. Where they overlap constructively, we see brightness. Where they interfere destructively, darkness appears. Understanding single-slit diffraction is fundamental in optics and helps in explaining various concepts like the resolving power of optical instruments.

Diffraction pattern

The diffraction pattern generated by a single slit is a classic example of light displaying wave-like behavior. These patterns are characterized by the central maximum and numerous surrounding minima and maxima. The central part of the pattern is the brightest and broadest, known as the central maximum. Around it are a series of smaller and dimmer spots. These are the result of light waves bending around the edges of the slit and interfering with one another.

The position of these dark and bright fringes depends on the wavelength of the light and the width of the slit. For instance, the condition for the first minima occurs when the path difference between light from the edges of the slit is exactly one wavelength. This is described by the formula: \( a \sin \theta = m \lambda \), where \( a \) is the slit width, \( \theta \) is the angle from the center of the pattern, \( m \) is the order of the minima, and \( \lambda \) is the wavelength. This equation helps determine the angles where darkness appears, thus mapping out the diffraction pattern.

Wavelength of light

The wavelength of light, denoted by \( \lambda \), is a critical factor in diffraction phenomena. It represents the distance between two successive peaks of a wave. In the context of diffraction, the wavelength determines how and where the light will spread out and interfere after passing through a slit.

For light of wavelength 580 nm, as given in the example, the diffraction pattern's features—such as how the light and dark bands (minima and maxima) are spaced—are also dictated by this wavelength. A longer wavelength results in a more spread-out pattern, with bands further apart. A shorter wavelength causes a more compressed pattern.

Understanding the wavelength's role helps in predicting and analyzing optical phenomena. It also aids in designing experiments and interpreting the results. Wavelength is not only crucial for visible light but also for understanding various applications ranging from telecommunications to spectroscopy.

Intensity distribution

Intensity distribution refers to how the light's brightness varies across the diffraction pattern. In single-slit diffraction, the intensity, or brightness, of light does not remain constant but changes depending on the angle \( \theta \) relative to the central maximum.

For a single slit, the intensity at any angle \( \theta \) is given by the formula: \( I(\theta) = I_0 \left(\frac{\sin(\beta)}{\beta}\right)^2 \), where \( \beta = \frac{\pi a \sin \theta}{\lambda} \) and \( I_0 \) is the maximum intensity at \( \theta = 0 \). This means that the light is brightest at the center and gradually diminishes as you move away, forming the distinct pattern of peaks and troughs.

At specific angles such as 45 degrees, the formula allows us to calculate the exact intensity relative to the central maximum. For example, in the provided exercise, at \( \theta = 45.0^\circ \), the intensity is about 40.5% of the center. These calculations are crucial for applications that rely on precise control and manipulation of light, such as optical engineering and beam shaping.