Problem 16 - University Physics with Modern Physics

Question

Monochromatic light of wavelength 592 nm from a distant source passes through a slit that is 0.0290 mm wide. In the resulting diffraction pattern, the intensity at the center of the central maximum \((\theta = 0^\circ)\) is 4.00 \(\times\) 10\(^{-5}\) W/m\(^2\). What is the intensity at a point on the screen that corresponds to \(\theta\) = 1.20\(^\circ\)?

Step-by-Step Solution

Answer

The intensity at \( \theta = 1.20^\circ \) is approximately \( 1.33 \times 10^{-8} \text{ W/m}^2 \).

Step 1: Understand the Diffraction Equation

The intensity of light in a single-slit diffraction pattern is given by the equation: \[ I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \]where \( I_0 \) is the intensity at the center, \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), and \( a \) is the slit width. We are tasked with finding \( I(1.20^\circ) \).

Step 2: Convert Units and Calculate \( \beta \)

Convert wavelength \( \lambda = 592 \text{ nm} = 592 \times 10^{-9} \text{ m} \) and slit width \( a = 0.029 \text{ mm} = 0.029 \times 10^{-3} \text{ m} \). For \( \theta = 1.20^\circ \), \( \sin(\theta) \approx \sin(1.20^\circ) = 0.0209 \). Calculate:\[ \beta = \frac{\pi \times 0.029 \times 10^{-3} \times 0.0209}{592 \times 10^{-9}} \]

Step 3: Simplify and Insert Values

Calculate \( \beta \) using the given values:\[ \beta = \frac{\pi \times 0.029 \times 10^{-3} \times 0.0209}{592 \times 10^{-9}} \approx 3.20 \]This will be used to find the new intensity at \( \theta = 1.20^\circ \).

Step 4: Compute the Intensity Ratio

Using the intensity equation:\[ \frac{I(\theta)}{I_0} = \left( \frac{\sin(3.20)}{3.20} \right)^2 \]Calculate \( \sin(3.20) \) and divide by \( 3.20 \). Then square the result to find the intensity ratio.

Step 5: Determine the Intensity at \( 1.20^\circ \)

Calculate \( \sin(3.20) \approx -0.0584 \) (note the negative value, but the square is taken):\[ \frac{I(1.20^\circ)}{4.00 \times 10^{-5}} = \left( \frac{-0.0584}{3.20} \right)^2 \approx \left( -0.01825 \right)^2 \approx 0.0003335 \]Therefore, \( I(1.20^\circ) = 4.00 \times 10^{-5} \times 0.0003335 \approx 1.33 \times 10^{-8} \text{ W/m}^2 \).

Key Concepts

Diffraction Pattern

When light encounters an obstacle, such as a slit, it tends to bend and spread out. This is known as diffraction. In a single-slit diffraction setup, when monochromatic light passes through a narrow slit, it produces a characteristic pattern of alternating bright and dark bands on a screen, known as a diffraction pattern.
This pattern results from the constructive and destructive interference of light waves. The central bright band, called the central maximum, is the brightest part of the pattern. Flanking it are alternating bands of light (maxima) and darkness (minima), whose intensity gradually decreases with distance from the center.

Intensity Calculation

The intensity of light in a diffraction pattern is not uniform. It varies based on the angle \(\theta\)\, which measures the position on the screen relative to the central maximum. The intensity at any point in the pattern can be calculated using:

• \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2\)

Where \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\)\, \(I_0\) is the intensity at the central maximum, and \(\lambda\) is the wavelength. This equation accounts for the wave nature of light and demonstrates how intensity diminishes as one moves away from the central axis. Calculating \(\beta\) provides insight into intensity changes due to different angles and slit dimensions.

Monochromatic Light

Monochromatic light is light consisting of a single wavelength and frequency. This type of light is crucial for diffraction experiments because it ensures clear and distinguishable patterns.
In the given exercise, the light used has a wavelength of 592 nm. As all photons have the same energy and wavelength, the pattern produced is coherent and precise, aiding in accurate determination of diffraction characteristics such as maxima and minima positions.

Wavelength

The wavelength \(\lambda\) of light is the distance between consecutive peaks (or troughs) in a wave. It’s a vital parameter in understanding many optical phenomena, including diffraction. The wavelength determines the scale of the diffraction pattern.

• Longer wavelengths create more spread-out patterns,
• Shorter wavelengths yield narrower bands.

In our context, the light's wavelength is 592 nm, translating to 592 \(\times\) 10^{-9}\ m. This small scale emphasizes the microscopic nature of diffraction patterns.

Slit Width

Slit width is the width (denoted as \(a\)) of the opening through which light passes. It significantly influences the diffraction pattern's appearance. As the width changes, so does the interference pattern's distribution:

• Narrower slits cause more pronounced diffraction, leading to wider separation between maxima.
• Wider slits result in less diffraction and narrower spacing of the bands.

In this exercise, the slit width is 0.0290 mm, which defines the initial condition for the diffraction phenomena observed.