Problem 13 - University Physics with Modern Physics

Question

Two very narrow slits are spaced 1.80 \(\mu\)m apart and are placed 35.0 cm from a screen. What is the distance between the first and second dark lines of the interference pattern when the slits are illuminated with coherent light with \(\lambda\) = 550 nm? (Hint: The angle \(\theta\) in Eq. (35.5) is \(not\) small.)

Step-by-Step Solution

Answer

The distance between the first and second dark lines is approximately 3.1 mm.

Step 1: Understand the Problem

We need to calculate the distance between the first and second dark lines in the interference pattern from two slits using the given wavelength \(\lambda = 550\ \text{nm}\), slit separation \(d = 1.80\ \mu\text{m}\), and distance to the screen \(L = 35\ \text{cm}\). The dark lines appear where destructive interference occurs, given by the path difference \(d\sin\theta = \left(m + \frac{1}{2}\right)\lambda\), where \(m\) is an integer for dark lines.

Step 2: Use the Condition for Dark Lines

For the first dark line, \(m = 0\) so the condition becomes:\[d\sin\theta_1 = \left(0 + \frac{1}{2}\right)\lambda = \frac{1}{2}\lambda\]For the second dark line, \(m = 1\) so:\[d\sin\theta_2 = \left(1 + \frac{1}{2}\right)\lambda = \frac{3}{2}\lambda\]

Step 3: Calculate the Angles \(\theta_1\) and \(\theta_2\)

Solve for \(\theta_1\) and \(\theta_2\) using the equations:\[\sin\theta_1 = \frac{\frac{1}{2}\lambda}{d} = \frac{275}{1800}\approx 0.1528\]\[\sin\theta_2 = \frac{\frac{3}{2}\lambda}{d} = \frac{825}{1800}\approx 0.4583\]Find \(\theta_1 = \arcsin(0.1528)\) and \(\theta_2 = \arcsin(0.4583)\).

Step 4: Calculate Position of Dark Lines on the Screen

Use \(y = L\tan\theta\) to find the position \(y_1\) for the first dark line and \(y_2\) for the second:\[y_1 = 35\times\tan(\theta_1)\]\[y_2 = 35\times\tan(\theta_2)\]Compute \(y_1\) and \(y_2\).

Step 5: Find the Distance Between the Dark Lines

The distance \(\Delta y\) between the first and second dark lines is:\[\Delta y = y_2 - y_1\]Calculate the difference between the two positions obtained in Step 4 to find \(\Delta y\).

Key Concepts

Destructive interference

When we talk about interference patterns, "destructive interference" is a crucial concept. This happens when two or more waves overlap in such a way that they cancel each other out. The high points of one wave meet the low points of another, resulting in reduced or zero amplitude. This effect is very easy to observe in light waves, with the result being areas that appear dark on a screen.

In the context of our double-slit experiment, destructive interference results in dark lines on the screen where the waves from the two slits perfectly cancel each other out. This occurs when the path difference between the two waves is a half-multiple of the wavelength, specifically \[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]where:

• \(d\) is the distance between the slits,
• \(\lambda\) is the wavelength of the light,
• \(\theta\) is the angle of the dark fringe from the horizontal,
• \(m\) is an integer (0, 1, 2, ...).

This formula helps us find where the darkness happens, indicating complete wave cancellation on our pattern.

Double-slit experiment

The double-slit experiment, first conducted by Thomas Young, is a famous demonstration of the wave properties of light. In this experiment, coherent light — light of a single wavelength — shines on a pair of closely-spaced slits. As the light passes through these slits, it spreads out or diffracts.

The resulting light waves overlap and interfere with each other, producing distinct patterns on a screen behind the slits. We see a series of bright and dark lines. Bright lines occur from constructive interference, while dark lines, where no light appears, result from destructive interference.

This experiment is not only foundational in understanding wave properties of light but also delves into the principles of superposition, where the resulting wave pattern is a sum of individual wave patterns. As in our problem, by knowing slit spacing and wavelength, interference patterns teach us about wave interactions on a detailed level.

Wavelength calculation

Wavelength calculation is an integral part of solving interference problems. Wavelength, denoted as \(\lambda\), measures the distance between consecutive peaks of a wave. It's a key parameter that determines where the interference pattern's bright and dark spots will appear.

In calculating where dark lines form in interference patterns (as in our example), we use the formula specific to destructive interference:\[ d\sin\theta = \left(m + \frac{1}{2}\right)\lambda \]Here, \(m\) is the order of the dark fringe, and knowing the position \(\theta\) helps us locate these dark spots precisely on the screen.

To find where these lines appear on the screen, converting angles to linear position using \( y = L\tan\theta \), where \(L\) is the distance from the slits to the screen, is crucial. All these calculations revolve around understanding wavelength, showing its central role in predicting wave interference patterns effectively.