Question

Two light sources can be adjusted to emit monochromatic light of any visible wavelength. The two sources are coherent, 2.04 \(\mu\)m apart, and in line with an observer, so that one source is 2.04 \(\mu\)m farther from the observer than the other. (a) For what visible wavelengths (380 to 750 nm) will the observer see the brightest light, owing to constructive interference? (b) How would your answers to part (a) be affected if the two sources were not in line with the observer, but were still arranged so that one source is 2.04 \(\mu\)m farther away from the observer than the other? (c) For what visible wavelengths will there be \(destructive\) interference at the location of the observer?

Short answer

(a) For constructive interference: 680 nm, 510 nm, etc. (b) Same results apply if not inline but equidistant changes. (c) For destructive interference: 582 nm, etc.

Step-by-step solution

Understanding Constructive Interference

For constructive interference to occur, the path difference between the light from the two sources must be an integer multiple of the wavelength. This condition can be expressed with the equation: \[ \Delta d = m \lambda \] where \( \Delta d \) is the path difference, \( m \) is an integer (0, 1, 2, ...), and \( \lambda \) is the wavelength. Given that \( \Delta d = 2.04 \mu m \), we can find \( \lambda \) for different values of \( m \).

Calculating Wavelengths for Constructive Interference

We rearrange the equation to find the wavelength: \[ \lambda = \frac{\Delta d}{m} = \frac{2.04 \mu m}{m} \] Convert \( 2.04 \mu m \) to nm: \( 2.04 \mu m = 2040 \) nm. We seek wavelengths in the range of 380 nm to 750 nm. Solve for \( m \) by checking the integer values that keep \( \lambda \) within this range: \( m = 3, 4, 5, \) and so on until those conditions are met.

Solving for Constructive Interference Wavelengths

Calculate \( \lambda \) for each integer \( m \): - For \( m = 3 \), \( \lambda = \frac{2040}{3} = 680 \) nm (valid)- For \( m = 4 \), \( \lambda = \frac{2040}{4} = 510 \) nm (valid)- Check larger values to see if they fit within the visible range. These calculated wavelengths ensure constructive interference occurs.

Analyzing Effect if Sources Are Not Inline

If the sources are not inline but maintain the distance such that one source is 2.04 \( \mu m \) farther away, the condition for constructive interference doesn't change. Thus, the wavelengths calculated remain the same.

Understanding Destructive Interference

For destructive interference, the path difference is an odd half multiple of the wavelength. The condition for this is given by: \[ \Delta d = \left( m + \frac{1}{2} \right) \lambda \]. Using similar conversion and path difference \( 2040 \) nm, solve for \( \lambda \) within the visible range.

Calculating Wavelengths for Destructive Interference

The wavelengths can be calculated using \[ \lambda = \frac{2040}{m + \frac{1}{2}} \] ensuring \( \lambda \) is between 380 nm to 750 nm. Solve for integer \( m \) with values such as \( m = 2, 3, 4, \) etc., similar to the constructive interference.

Solving for Destructive Interference Wavelengths

With the formula, calculate for different \( m \):- For \( m = 2 \), \( \lambda = \frac{2040}{2.5} = 816 \) nm (not visible)- For \( m = 3 \), \( \lambda = \frac{2040}{3.5} = 582 \) nm (valid)- Find all \( \lambda \) within those bounds for visible light. It reveals the wavelengths where destructive interference occurs.

Key concepts

Constructive Interference

When we talk about light and interference, constructive interference refers to the situation where two light waves meet and combine to produce a wave of greater amplitude. This leads to brighter light observations. For constructive interference to occur, the path difference between waves from two sources must be an integer multiple of their wavelength.

To calculate the specific wavelengths at which constructive interference happens, use the formula:
\[ \Delta d = m \lambda \]Here, \( \Delta d \) is the path difference, \( m \) is an integer, and \( \lambda \) is the wavelength of the light.

By substituting the given path difference of 2040 nm, you can determine valid wavelengths of light by checking for integer values of \( m \) which provide wavelengths within the visible range of 380 nm to 750 nm. Some valid wavelengths include 680 nm for \( m = 3 \) and 510 nm for \( m = 4 \). What changes if the arrangement of light sources is altered but maintains the path difference is that the conditions for constructive interference still apply. Thus, the calculated wavelengths remain valid regardless of the positioning of sources relative to the observer.

Destructive Interference

Contrasting with constructive interference, destructive interference is when two waves meet and cancel each other out, resulting in a wave of reduced or no amplitude. This leads to dimmer light or darkness. Destructive interference occurs when the path difference is an odd multiple of half the wavelength:

\[\Delta d = \left( m + \frac{1}{2} \right) \lambda\]This equation means that waves are out of phase, creating cancellation.

To find visible wavelengths that lead to destructive interference, solve the equation by substituting the path difference of 2040 nm and varying \( m \). The condition will only be satisfied for certain wavelengths that fall within the visible light spectrum of 380 nm to 750 nm. For instance, a wavelength of 582 nm occurs for \( m = 3 \), demonstrating how specific values of \( \lambda \) result in destructive interference.
Keep in mind that while calculating, the wavelengths must fall within the visible spectrum to observe them. The math reveals how and why certain wavelengths become 'invisible' to an observer due to destructive interference.

Visible Wavelengths

Visible wavelengths are the portion of the electromagnetic spectrum that can be detected by the human eye. These wavelengths range from about 380 nm (violet) to 750 nm (red). The colors we perceive depend on the wavelength of light. Shorter wavelengths are closer to violet and blue, while longer wavelengths shift towards red.

When discussing interference, the visibility of light can dramatically shift due to effects like constructive and destructive interference. Constructive interference can make specific wavelengths stand out more brightly, while destructive interference can cause them to 'disappear'.
To calculate which wavelengths are affected, it's critical to understand that the interaction between light waves and their relative phase plays a pivotal role in what the observer sees. For instance, when the wavelength corresponds to constructive interference conditions, that wavelength will appear more vibrant, whereas destructive interference can completely negate the visibility of a particular wavelength.
Understanding visible wavelengths is key to predicting and explaining the patterns and colors observed when light waves interact, whether through diffraction or interference. This knowledge helps elucidate how natural phenomena like rainbows or soap bubbles get their vivid colors from specific interference patterns.