Question

A Young's interference experiment is performed with monochromatic green light \((\lambda=540 \mathrm{nm}) .\) The separation between the slits is \(0.100 \mathrm{~mm},\) and the interference pattern on a screen shows the first side maximum \(5.40 \mathrm{~mm}\) from the center of the pattern. How far away from the slits is the screen?

Short answer

Answer: The distance between the slits and the screen is approximately 1.0 m.

Step-by-step solution

Write the formula for the position of the maxima in an interference pattern

Write the formula as \(y_n = \frac{n\lambda L}{d}\).

Plug in the given values

We know the following values: - Wavelength \(\lambda = 540 \,\text{nm} = 540 \times 10^{-9}\, \text{m}\) (converting to meters) - Separation between the slits \(d = 0.100\, \text{mm} = 0.100 \times 10^{-3} \,\text{m}\) (converting to meters) - Distance between the first side maximum and the center of the pattern \(y_1 = 5.40\, \text{mm} = 5.40 \times 10^{-3}\, \text{m}\) (converting to meters) - Order of the maximum \(n = 1\) Now plug these values into the formula: \(5.40\times 10^{-3}=\frac{1\times 540\times 10^{-9} L}{0.100\times 10^{-3}}\)

Solve for L

To find the value of \(L\), we will first multiply both sides of the equation by \(0.100\times 10^{-3}\): \(5.40\times 10^{-3} \cdot 0.100 \times 10^{-3} = 540 \times 10^{-9} L\) Now, divide both sides by the remaining factor on the right side of the equation, which in this case is \(540 \times 10^{-9}\): \(L = \frac{5.40 \times 10^{-3} \cdot 0.100\times 10^{-3}}{540\times 10^{-9}}\) Upon calculating, we get: \(L \approx 1.0\,\text{m}\) Thus, the distance between the slits and the screen is approximately 1 meter.

Key concepts

Monochromatic Light

Monochromatic light refers to light that consists of one single wavelength. In the context of Young's interference experiment, using monochromatic light allows for a clearer and more distinct interference pattern. This is because all the light waves have the same frequency and phase, which leads to consistent constructive and destructive interference effects. For example, in the exercise above, the monochromatic light has a wavelength of 540 nm, which corresponds to green light.

Monochromatic light can be produced by sources such as lasers or specialized lamps. Using monochromatic light simplifies calculations in interference experiments, as variations in wavelength can introduce complex variables. This consistency is crucial for clarity and accuracy in analyzing interference patterns.

Interference Pattern

An interference pattern emerges when waves, such as light waves, overlap and interact with each other. In Young's double-slit experiment, waves passing through two slits converge on a screen and either amplify or cancel each other out, depending on their phase difference. This results in a series of bright and dark fringes.

The bright bands in the pattern are due to constructive interference, where the waves are in phase and augment each other. Conversely, the dark bands result from destructive interference, where waves are out of phase and cancel each other. The pattern's appearance provides valuable insights into the wavelengths of the light and the nature of wave interaction.

For students, understanding this pattern is crucial for predicting how waves will behave in various conditions, aiding in more complex wave-related studies.

Wavelength Calculation

Calculating wavelength is a fundamental step in Young's interference experiment. It involves understanding the relationship between the observed interference pattern and the wavelength of the light used. The formula for the position of the maxima, \[y_n = \frac{n\lambda L}{d},\] connects the wavelength \( \lambda \) to the variables involved in the experiment.

In this case, knowing the distance of the maxima from the center, the order of maxima \(n)\), slit separation \(d)\), and the screen distance \(L)\) allows us to back-solve for any one of these when the wavelength is known. This relationship underscores the powerful role wavelength plays in determining the nature of the interference pattern.

Mastering wavelength calculations helps students make precise predictions in experiments involving light, sound, and other wave-based phenomena.

Slit Separation

Slit separation \(d)\) is a critical factor in Young's interference experiment. It determines the spacing between the two slits through which light passes. The relationship between slit separation and the interference pattern can be observed through the formula used to find maxima positions.

• If the slit separation is increased, the fringes on the screen get closer together.
• Conversely, smaller slit separation results in wider spacing between the fringes.

Understanding this relationship helps to adjust the experiment for desired results and finer measurements, like determining specific wavelength values.

In practical terms, knowing about slit separation allows for better control over experimental setups, enhancing the accuracy of detected interference patterns and the validity of the resulting data.