Question

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_0\) . (a) At a point in the pattern where the phase difference between the waves from the two slits is 60.0\(^\circ\), what is the intensity? (b) What is the path difference for 480-nm light from the two slits at a point where the phase difference is 60.0\(^\circ\)?

Short answer

(a) \( \frac{3}{4} I_0 \); (b) 40 nm.

Step-by-step solution

Understanding Interference Intensity

In a two-slit interference experiment, the intensity at any point on the screen is related to the phase difference \( \Delta \phi \) between the light waves coming from the two slits. The formula for intensity \( I \) at a point where the phase difference \( \Delta \phi \) is given by \[ I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right) \] where \( I_0 \) is the intensity at the central maximum.

Calculate Intensity for Phase Difference 60°

Given \( \Delta \phi = 60.0^\circ \), convert this into radians as most trigonometric calculations are done in radians. \( \Delta \phi = \frac{\pi}{3} \) radians. Substitute \( \Delta \phi \) into the formula: \[ I = I_0 \cos^2\left(\frac{\pi}{6}\right) \] Calculate \( \cos\left(\frac{\pi}{6}\right) \), which equals \( \frac{\sqrt{3}}{2} \). Thus: \[ I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4} \] Hence, \( I = \frac{3}{4}I_0. \)

Understanding Path Difference

The phase difference \( \Delta \phi \) is related to the path difference \( \Delta x \) by the formula \[ \Delta \phi = \frac{2\pi \Delta x}{\lambda} \] where \( \lambda \) is the wavelength of light.

Calculate Path Difference for Phase Difference 60°

Given \( \Delta \phi = \frac{\pi}{3} \) (for 60° phase difference) and \( \lambda = 480 \) nm, substitute into the equation \( \Delta \phi = \frac{2\pi \Delta x}{\lambda} \): \[ \frac{\pi}{3} = \frac{2\pi \Delta x}{480} \] Simplify and solve for \( \Delta x \): \[ \Delta x = \frac{480}{2\cdot 3} = 40 \text{ nm} \] Therefore, the path difference is 40 nm.

Key concepts

Interference Pattern

In a two-slit interference setup, the term "interference pattern" refers to the series of bright and dark bands or spots that appear on a screen. These bands arise when coherent light waves, emanating from two narrow slits, superpose and interfere with each other.
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When waves overlap, they either reinforce each other, called constructive interference, leading to brighter bands, or they cancel each other out, called destructive interference, resulting in darker regions. This beautiful pattern is a physical demonstration of the wave nature of light.
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Key characteristics of the interference pattern include:
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• The spacing between the bright and dark bands depends on the wavelength of the light and the distance between the slits.


• The central bright band, also called the central maximum, is the most intense due to maximum constructive interference.

Intensity Calculation

The intensity of light in a two-slit interference pattern changes depending on where you observe it. This intensity, denoted as \( I \), at any point on the screen is linked to how the phase of the two light waves differs at that point.
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To calculate this, we use the formula:\[I = I_0 \cos^2\left(\frac{\Delta \phi}{2}\right)\]Here, \( I_0 \) represents the intensity at the peak of the central maximum, where phase difference is zero.
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• At a phase difference of 60.0\(^\circ\), converting to radians gives \( \frac{\pi}{3} \).


• Using this, intensity can be computed as \( I = \frac{3}{4}I_0 \).


• Thus, the intensity at this point is 75% of the central peak intensity.

Phase Difference

In the context of two-slit interference, phase difference \( \Delta \phi \) is the difference in phase between the wavefronts arriving from the two slits at a particular point on the observation screen.
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This phase difference is an angle typically measured in degrees or radians, where 360 degrees (or \( 2\pi \) radians) represent a full cycle.
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• The phase difference determines whether the interference at that point is constructive or destructive.


• Constructive interference, resulting in maximum intensity, occurs when \( \Delta \phi \) is a multiple of 2π.


• Destructive interference occurs when \( \Delta \phi \) is an odd multiple of \( \pi \).

Path Difference

Path difference, denoted as \( \Delta x \), is the difference in the distance traveled by two waves from the slits to a point on the screen.
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This concept is crucial as the path difference leads to a phase difference between the waves, thereby influencing the interference pattern.
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To determine the path difference corresponding to a specific phase difference, we use the relation:\[\Delta \phi = \frac{2\pi \Delta x}{\lambda}\]Where \( \lambda \) is the wavelength of the light used.
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• For a phase difference of 60.0\(^\circ\) or \( \frac{\pi}{3} \), the path difference would be 40 nm for 480-nm light.


• This path difference of 40 nm causes the waves to interfere at a different intensity than at the central maximum.

Wavelength

Wavelength, symbolized as \( \lambda \), is the distance over which the wave's shape repeats. It is a fundamental property of waves, including light, and is typically measured in nanometers (nm) for light waves.
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The wavelength determines the color of the light and the spacing of the interference pattern formed.
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• A smaller wavelength results in closer fringes in the interference pattern.


• In this context, we are using a light wavelength of 480 nm, which is characteristic of blue and violet light, contributing to the distinctive pattern observed.


• Wavelength also plays a critical role in calculating both path and phase differences.