Question

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25\(^\circ\) from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_0\)?

Short answer

(a) Using the formula with given values, \(\lambda = 365.4 \text{ nm}\). (b) Intensity \(I(\theta) = 0.19 I_0\).

Step-by-step solution

Relate phase difference to path difference

The path difference across the slit is related to the phase difference by the formula:\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \], where \(\Delta \phi\) is the phase difference, \(\lambda\) is the wavelength, and \(\Delta x\) is the path difference across the slit. We are given that \(\Delta \phi = 56.0\, \text{rad}\).

Calculate path difference using angle

The path difference \(\Delta x\) for a point \(\theta\) from the central maximum is given by \(\Delta x = a \sin(\theta)\), where \(a\) is the slit width (0.105 mm). The angle \(\theta\) is given as 3.25° or 0.0567 radians (after converting from degrees to radians).

Substitute values to find wavelength

Substitute the values into the phase difference equation to find the wavelength: \(56.0 = \frac{2\pi}{\lambda} \times 0.105 \times 10^{-3} \times \sin(0.0567)\). Solve for \(\lambda\) to find the wavelength.

Calculate the intensity using the Fraunhofer diffraction formula

The intensity \(I(\theta)\) at a given angle \(\theta\) compared to the central maximum \(I_0\) is given by the formula: \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \), where \(\beta = \frac{\pi a \sin(\theta)}{\lambda}\).

Substitute wavelength to find intensity at the point

Use the previously calculated wavelength and substitute it in the expression for \(\beta\). Then calculate: \( \beta = \frac{\pi \times 0.105 \times 10^{-3} \times \sin(0.0567)}{\lambda}\). Find \(\sin(\beta)\) and use the Fraunhofer equation to find the intensity \(I(\theta)\).

Key concepts

Monochromatic Radiation

Monochromatic radiation refers to light that consists of a single wavelength. This concept is essential in understanding single-slit diffraction patterns, as the light source's uniformity directly impacts the pattern's clarity and predictability. This kind of radiation is often used in experiments because it creates distinct, measurable patterns. Monochromatic sources, like lasers, ensure that the resulting diffraction pattern is clean and uncomplicated by multiple wavelengths interacting and overlapping.
This simplicity allows for easier calculations and a deeper understanding of the wave properties involved. In a single-slit experiment, knowing that the light is monochromatic helps us predict and calculate the pattern using precise mathematical models.

Phase Difference

In physics, the phase difference between waves refers to the amount by which one wave lags or leads another. It is measured in radians, where a complete cycle corresponds to a phase difference of 2\( \pi \) radians. In the context of single-slit diffraction, the phase difference is crucial for understanding how waves arriving from different parts of the slit interfere with each other.

• The formula \( \Delta \phi = \frac{2\pi}{\lambda} \Delta x \) relates phase difference \( \Delta \phi \) to wavelength \( \lambda \) and path difference \( \Delta x \).
• A larger phase difference results in more significant interference effects, influencing the position and size of dark and bright fringes in the diffraction pattern.

Understanding phase differences allows for predicting and calculating the fringe pattern in detail, giving insights into factors like light frequency and slit width.

Path Difference

Path difference is a term that describes the difference in distance traveled by two waves from their origin to a common point of observation. In the single-slit diffraction pattern, the path difference determines how the waves interfere when they merge on a surface.
The relationship between path difference and the diffraction pattern is governed by the equation \( \Delta x = a \sin(\theta) \), where \( a \) is the width of the slit and \( \theta \) is the angle from the central maximum. Calculating the path difference helps in understanding whether constructive or destructive interference occurs at a particular point in the diffraction pattern.
This understanding is critical in predicting where bright and dark bands will form, based on the geometry and properties of the light and slit.

Fraunhofer Diffraction Formula

The Fraunhofer diffraction formula is a paramount tool for calculating the intensity patterns in single-slit diffraction. It helps to quantify how light intensity varies at different angles from the slit.

• The formula \( I(\theta) = I_0 \left( \frac{\sin(\beta)}{\beta} \right)^2 \) relates the intensity \( I(\theta) \) of light at an angle \( \theta \) to the maximum intensity \( I_0 \).
• The parameter \( \beta \) is given by \( \beta = \frac{\pi a \sin(\theta)}{\lambda} \), where \( a \) is the slit width and \( \lambda \) is the wavelength.

This formula empowers physicists to predict the precise distribution of light and dark bands in the diffraction pattern, providing detailed insights into wave behaviors, based on fundamental wave properties. By employing the Fraunhofer formula, scientists and engineers can accurately model light behavior, enhancing applications ranging from optical instrument design to precise scientific measurements.