Question

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (\(\theta\) = 0\(^\circ\)) is \(I_0\) . What is the distance on the screen from the center of the central maximum (a) to the first minimum; (b) to the point where the intensity has fallen to \(I_0\) /2?

Short answer

(a) The distance to the first minimum is about 1.14 mm; (b) the distance to the point where intensity is \(I_0/2\) is about 0.57 mm.

Step-by-step solution

Understanding the Problem

We have two slits spaced 0.260 mm apart, and they are 0.900 m away from a screen. The coherent light used has a wavelength of 660 nm. We need to find two things: first, the distance to the first minimum, and second, the distance to the point where the intensity is half of the maximum. Let's solve the first part first.

Calculating the Position of the First Minimum

For the position of the first minimum in a double-slit interference pattern, we use the equation for destructive interference: \( d \sin\theta = (m + \frac{1}{2})\lambda \), where \( m \) is the order of the minimum. For the first minimum, \( m = 0 \). Hence, \( d \sin\theta = (0.5)\lambda \). Convert the distance d into meters and \( \lambda \) into meters. Then, solve for \( \sin\theta \). Use small angle approximation \( \sin \theta \approx \theta \approx \tan \theta \). The distance \( y \) on the screen is \( y = L \tan \theta \approx L \sin \theta \).

Converting Units and Solving for \(\theta\)

Convert given values: \(d = 0.260 \text{ mm} = 0.260 \times 10^{-3} \text{ m}\) and \(\lambda = 660 \text{ nm} = 660 \times 10^{-9} \text{ m}\). Plug these values into the equation: \(0.260 \times 10^{-3} \sin \theta = 0.5 \times 660 \times 10^{-9}\). Solve for \(\sin \theta\) and approximate \(\theta \approx \sin \theta\).

Finding the Distance from the Center

Calculate the distance \(y\) using \(y = L \sin \theta\), where \( L = 0.900 \text{ m} \). Simplify to find \( y \). This distance \( y \) will be the distance from the center of the central maximum to the first minimum on the screen.

Intensity at \( I_0 /2 \)

The intensity for a double-slit configuration is given by \( I = I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda} \right)\). Set \( I = \frac{I_0}{2} \) and solve for \(\theta\) using \( \cos^2\left(\frac{\pi d \sin\theta}{\lambda} \right) = \frac{1}{2} \), which simplifies to \(\theta = \frac{\lambda}{4d}\). Use small angle approximation \(\tan \theta \approx \sin \theta \).

Calculating \( \theta \) for \( I_0/2 \)

Set \(\theta = \frac{660 \times 10^{-9}}{4 \times 0.260 \times 10^{-3}}\). Convert this equation to find \(\theta\). Apply \(y = L \sin \theta\approx L \theta\) to find the position on the screen.

Key concepts

Destructive Interference

In a double-slit interference experiment, destructive interference occurs when waves from the two slits arrive at a point on the screen out of phase, thus canceling each other out. This leads to a point of minimum intensity. To find these points, we use the formula for destructive interference: - For destructive interference: \( d \sin \theta = (m + \frac{1}{2}) \lambda \)Here, \( d \) is the distance between the slits, \( \theta \) is the angle relative to the central maximum, \( m \) is the order of the minimum (an integer starting from 0), and \( \lambda \) is the wavelength of the light. For the first minimum, \( m = 0 \), simplifying this equation to: - \( d \sin \theta = \frac{1}{2} \lambda \).This condition ensures that the difference in path length between the two waves is half a wavelength, leading to complete cancellation at that point. Using the small angle approximation of \( \sin \theta \approx \theta \approx \tan \theta \), the position of this first minimum on the screen can be calculated.

Coherent Light

Coherent light is essential in interference experiments like the double-slit experiment. Coherent light beams maintain a constant phase difference and have the same wavelength. Their uniform phase relationship allows the light from each slit to produce a predictable and stable interference pattern on the screen. Key characteristics of coherent light: - It arises from a single source, or synchronized sources, ensuring the same frequency and waveform. - It's crucial for observing clear and consistent interference patterns in experiments. - Continuous phase relationships minimize changes in the constructive and destructive interference locations. In this exercise, light with a wavelength of 660 nm is used, producing visible and stable interference patterns that include regions of maximum and minimum intensity.

Intensity Pattern

The intensity pattern in a double-slit interference experiment is characterized by alternating bright and dark fringes on the screen. The bright fringes correspond to constructive interference, where the path difference between light from the two slits is a multiple of the wavelength. Conversely, dark fringes result from destructive interference. For intensity: - Intensity for constructive interference is maximized at the central maximum.- The formula for intensity is \( I = I_0 \cos^2\left(\frac{\pi d \sin\theta}{\lambda} \right) \), where \( I_0 \) is the peak intensity.When considering the intensity at half the maximum: - Set \( I = \frac{I_0}{2} \) and solve the equation \( \cos^2\left(\frac{\pi d \sin\theta}{\lambda} \right) = \frac{1}{2} \).- This yields points for halfway diminished intensity, outlining intermediate positions between bright and dark fringes.

First Minimum Position

The first minimum position in a double-slit interference pattern is the location on the screen where the first dark fringe appears. This corresponds to the lowest intensity point next to the central bright fringe. To determine this position, use the equation: - \( y = L \tan \theta \approx L \sin \theta \), with \( L \) being the distance from the slits to the screen.In this experiment:- Convert the slit separation and wavelength into meters: \( d = 0.260 \times 10^{-3} \text{ m} \) and \( \lambda = 660 \times 10^{-9} \text{ m} \).- Use the equation for destructive interference: \( 0.260 \times 10^{-3} \sin \theta = 0.5 \times 660 \times 10^{-9} \).Solving for \( \theta \), you determine the angle where the first dark fringe appears. Applying \( y = 0.900 \times \sin \theta \), you find the actual distance of this fringe on the screen. This indicates where the contrast between bright and dark fringes begins, demonstrating the wave nature of light. Understanding the positioning of these fringes enriches comprehension of wave interference and the impact of variables such as wavelength and slit separation.