Question

Parallel rays of monochromatic light with wavelength 568 nm illuminate two identical slits and produce an interference pattern on a screen that is 75.0 cm from the slits. The centers of the slits are 0.640 mm apart and the width of each slit is 0.434 mm. If the intensity at the center of the central maximum is 5.00 \(\times\) 10\(^{-4}\) W/m\(^2\), what is the intensity at a point on the screen that is 0.900 mm from the center of the central maximum?

Short answer

Use the double-slit interference formula to calculate intensity: \(I = I_0 \cos^2(\Phi/2)\).

Step-by-step solution

Understand the Problem Setup

We aim to find the intensity at a point 0.900 mm from the center of the central maximum on the interference pattern created by two slits. We know the light wavelength (\(\lambda = 568\, \text{nm}\)), slit separation (\(d = 0.640\, \text{mm}\)), slit width (\(a = 0.434\, \text{mm}\)), and screen distance from slits (\(L = 75.0\, \text{cm}\)). The central maximum intensity is given as \(I_0 = 5.00 \times 10^{-4} \text{ W/m}^2\). We will use the formula for intensity in a double-slit interference pattern considering the path difference and phase.

Use the Double-Slit Intensity Formula

The intensity at a distance \(y\) from the center of the central maximum on the screen is given by \(I = I_0 \cos^2(\Phi/2)\), where \( \Phi = \frac{2\pi}{\lambda} \cdot d \cdot \frac{y}{L} \). This formula takes into account the phase difference between light arriving at point \(y\).

Calculate the Phase Difference \(\Phi\)

Substitute the known values into the phase difference formula:\[\Phi = \frac{2\pi}{568 \times 10^{-9}} \cdot 0.640 \times 10^{-3} \cdot \frac{0.900 \times 10^{-3}}{0.750}.\]Calculate \(\Phi\) to find the phase difference.

Evaluate \(\cos^2(\Phi/2)\)

Once \(\Phi\) is known, calculate \(\Phi/2\) and then \(\cos^2(\Phi/2)\). This value will be used to determine the intensity at the point 0.900 mm away from the center.

Substitute Back into Intensity Formula

Now that we have \(\cos^2(\Phi/2)\), substitute this back into the intensity formula: \(I = I_0 \cos^2(\Phi/2)\). This will give the intensity at the specified point on the screen.

Key concepts

Monochromatic Light

Monochromatic light refers to light of a single wavelength or color. This type of light is ideal for experiments involving wave interference, such as the classic double-slit experiment. The word "monochromatic" comes from Greek roots meaning "one color," which is precisely what this kind of light represents: a single wavelength, without the mixing of multiple colors. In the given exercise, the light used has a wavelength of 568 nanometers. The consistency of the wavelength allows for predictable interference patterns, as every wave crest and trough is aligned uniformly.

Monochromatic light sources, like lasers, help create clear and stable patterns in interference experiments. By utilizing light of a singular wavelength, scientists and students can more easily observe phenomena like fringe patterns on a screen. Such precision is crucial in carefully controlled experiments where variable factors must be minimized.

Intensity Pattern

The intensity pattern on a screen resulting from double-slit interference consists of alternating bright and dark regions. This occurs due to constructive and destructive interference of the light waves passing through the slits. At the center of the screen, known as the central maximum, the light waves are perfectly in phase, producing a bright spot.

As you move away from the central maximum, the intensity of light decreases. The intensity at any given point is determined by the interference of light waves arriving in phase or out of phase. Constructive interference, where waves align crest to crest, results in bright bands, while destructive interference, where waves align trough to crest, results in dark bands.

• Central maximum: the brightest point.
• Minima: points of destructive interference causing darkness.
• Maxima: bright points formed by constructive interference.

Understanding and calculating the intensity at different points require knowledge of the phase shifts between the incoming light waves.

Phase Difference

Phase difference refers to the relative position of one wave crest to another and is crucial in understanding interference patterns. In the context of the double-slit experiment, the phase difference determines whether waves interfere constructively or destructively. It is given by the angle \(\Phi\), which is calculated using the expression:\[\Phi = \frac{2\pi}{\lambda} \cdot d \cdot \frac{y}{L}\]where \(\lambda\) is the wavelength, \(d\) is the slit separation, \(y\) is the distance from the central maximum, and \(L\) is the distance from the slits to the screen.

A phase difference of multiples of \(2\pi\) indicates constructive interference, resulting in brightness, while phase difference in odd multiples of \(\pi\) indicates destructive interference, leading to darkness. Calculating the phase difference helps in predicting the intensity and location of the interference fringes on the screen.

Wave Interference

Wave interference occurs when two or more light waves meet, resulting in a new wave pattern. This pattern is critical in understanding phenomena like the double-slit interference pattern. Two types of interference are possible: constructive and destructive.

Constructive interference enhances wave actions as peaks meet peaks, increasing the resultant wave's amplitude. This leads to brighter spots on the screen. Destructive interference, on the other hand, happens when peaks meet troughs, canceling each other out, resulting in darkness. The periodic alternation between these two interferences forms the pattern seen on the screen.

• Constructive interference: results in heightened amplitude.
• Destructive interference: results in diminished amplitude.

Interference depends on the wavelength and relative phases of the waves. Understanding wave interference helps predict how the light from two slits will combine to form patterns of light and dark on the screen.