Problem 22 - University Physics with Modern Physics

Question

Two slits spaced \(0.0720 \mathrm{~mm}\) apart are \(0.800 \mathrm{~m}\) from a screen. Coherent light of wavelength \(\lambda\) passes through the two slits. In their interference pattern on the screen, the distance from the center of the central maximum to the first minimum is \(3.00 \mathrm{~mm} .\) If the intensity at the peak of the central maximum is \(0.0600 \mathrm{~W} / \mathrm{m}^{2},\) what is the intensity at points on the screen that are (a) \(2.00 \mathrm{~mm}\) and (b) \(1.50 \mathrm{~mm}\) from the center of the central maximum?

Step-by-Step Solution

Answer

Intensity at 2 mm is \(0.0386 \text{ W/m}^2\) and at 1.5 mm is \(0.0481 \text{ W/m}^2\).

Step 1: Formula for Intensity

To find the intensity at a point on the screen, we start by noting that the intensity due to interference of two coherent sources is given by the formula: \[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]where \( I_0 \) is the peak intensity (at the central maximum), and \( \Delta \phi \) is the phase difference.

Step 2: Determine Phase Difference

The phase difference \( \Delta \phi \) is related to the path difference, which can be expressed in terms of the distance \( y \) from the center on the screen, the slit separation \( d \), and the distance to the screen \( L \). Using geometrical approximation, \( \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \), where \(\sin \theta = \frac{y}{L}\).

Step 3: Calculate Wavelength

First, calculate the wavelength \( \lambda \). The position of the first minimum is given by the path difference \( \frac{\lambda}{2} \). Thus, \( d \sin \theta = \frac{\lambda}{2} \), and since the small angle approximation implies \( \sin \theta \approx \frac{y_1}{L} \), we have:\[ d \frac{y_1}{L} = \frac{\lambda}{2} \]Solving for \( \lambda \), we get:\[ \lambda = \frac{2dy_1}{L} \]The known values are \( d = 0.0720 \text{ mm} = 7.20 \times 10^{-5} \text{ m} \), \( y_1 = 3.00 \text{ mm} = 3.00 \times 10^{-3} \text{ m} \), and \( L = 0.800 \text{ m} \). Substitute these to find \( \lambda \).

Step 4: Calculate Intensity at 2 mm

Substitute \( y = 2.00 \text{ mm} = 2.00 \times 10^{-3} \text{ m} \) in the path difference formula, calculate \( \Delta \phi \), then find intensity:1. \( \sin \theta = \frac{2.00 \times 10^{-3}}{0.800} \).2. Calculate \( \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \).3. Finally, calculate \( I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \) to find the intensity at \( y = 2 \text{ mm} \).

Step 5: Calculate Intensity at 1.5 mm

Substitute \( y = 1.50 \text{ mm} = 1.50 \times 10^{-3} \text{ m} \) into the same formulas:1. \( \sin \theta = \frac{1.50 \times 10^{-3}}{0.800} \).2. Calculate \( \Delta \phi = \frac{2\pi d \sin \theta}{\lambda} \).3. Finally, calculate \( I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \) to find the intensity at \( y = 1.5 \text{ mm} \).

Key Concepts

double-slit experiment

The double-slit experiment is a fundamental demonstration of the wave nature of light. It involves passing coherent light, such as a laser, through two closely spaced slits. As the light waves emerge from the slits, they overlap and create an interference pattern on a screen placed at some distance. This pattern consists of a series of bright and dark fringes or bands.
Several factors influence this pattern:

• Wavelength (\( \lambda \)): The distance between successive peaks of the light wave. Longer wavelengths lead to wider patterns.
• Slit Separation (\( d \)): Larger separations result in more closely spaced fringes.
• Distance to Screen (\( L \)): The farther the screen, the more spread out the pattern.

The central fringe is the brightest, known as the central maximum, flanked by alternating bright and dark areas. These bright spots occur where the waves from the two slits meet in phase, causing constructive interference, while the dark bands occur where destructive interference cancels the waves.

wave phase difference

The phase difference between waves is a crucial aspect of the interference pattern observed in the double-slit experiment. It refers to the difference in the position of the crests and troughs of the two waves as they meet at a point on the screen. The phase difference (\( \Delta \phi \)) determines whether the interference is constructive or destructive.
Constructive interference occurs when the phase difference is an integer multiple of \( 2\pi \). At these points, the waves are in phase, reinforcing each other to produce bright fringes. Conversely, destructive interference arises when the phase difference equals an odd multiple of \( \pi \), leading to a dark fringe as the waves are out of phase.
The phase difference can be directly related to path difference (the difference in distances traveled by two waves), slit separation, and the angle (\( \theta \)) to the point on the screen:

• Path Difference (\( \Delta x \)): Given by \( d \sin \theta \).
• Relationship to Phase Difference: \( \Delta \phi = \frac{2\pi \Delta x}{\lambda} \).

This gives us a practical way to calculate phase differences for specific positions on the interference pattern.

intensity calculation

Calculating the intensity of the interference pattern at any point on the screen involves understanding how the superimposed waves interact. The intensity (\( I \)) at a point is dependent on the square of the amplitude of the resultant wave formed by the overlap of individual waves.
From the step-by-step solution, the intensity formula is given by:\[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \]Where:

• \( I_0 \) is the peak intensity at the central maximum.
• \( \Delta \phi \) is the phase difference.

The formula shows that the intensity varies with the cosine squared of half the phase difference. Thus, wherever \( \Delta \phi \) leads the cosine squared function to peak, you find maxima (bright fringes), whereas minima (dark fringes) occur at zeros of the cosine function.
To find the intensity at specific points, such as 2.00 mm and 1.50 mm from the center, it is essential to calculate \( \Delta \phi \) using the given slit separation and screen distance, then use the intensity formula. This method reveals how the overlap of waves creates a varying pattern of light on the screen.