Problem 38 - University Physics with Modern Physics

Question

Monochromatic light with wavelength 490 nm passes through a circular aperture, and a diffraction pattern is observed on a screen that is 1.20 m from the aperture. If the distance on the screen between the first and second dark rings is 1.65 mm, what is the diameter of the aperture?

Step-by-Step Solution

Answer

The diameter of the aperture is approximately 0.438 mm.

Step 1: Identify Key Information

We have the wavelength of the light \( \lambda = 490 \text{ nm} \), the distance from the aperture to the screen \( L = 1.20 \text{ m} \), and the distance between the first and second dark rings \( \Delta y = 1.65 \text{ mm} \). We need to find the diameter of the aperture.

Step 2: Convert Units

Convert the wavelength and distance between rings to meters: \( \lambda = 490 \times 10^{-9} \text{ m} \) and \( \Delta y = 1.65 \times 10^{-3} \text{ m} \).

Step 3: Use Diffraction Formula

The formula for the angular position of the minima in a circular aperture diffraction pattern is \( \theta = 1.22\frac{\lambda}{D} \), where \( D \) is the diameter of the aperture. The spacing \( \Delta y \) relates to the angle \( \Delta \theta \) by \( \Delta y = L \Delta \theta \). Since this is between first and second minima, we have \( \Delta \theta = 1.22\left(\frac{\lambda}{D}\right) - 1.22\left(\frac{\lambda}{D}\right) \), leading to \( \Delta y = 1.22\frac{\lambda}{D}L \).

Step 4: Rearrange to Solve for D

Rearrange the formula \( \Delta y = 1.22\frac{\lambda L}{D} \) to solve for \( D \): \( D = 1.22\frac{\lambda L}{\Delta y} \).

Step 5: Plug in Numbers

Substitute the values: \( D = 1.22\frac{(490 \times 10^{-9} \text{ m})(1.20 \text{ m})}{1.65 \times 10^{-3} \text{ m}} \). Calculate \( D \).

Step 6: Compute Final Answer

Carrying out the calculation, we find that \( D \approx 438 \times 10^{-6} \text{ m} = 0.438 \text{ mm} \).

Key Concepts

Circular Aperture

A circular aperture refers to a small, round opening through which light can pass. Imagine it like the hole you punch in a sheet of paper. When light travels through this hole, it creates interesting patterns due to the wave nature of light.

The significance of a circular aperture in physics often comes up when studying diffraction patterns. This is because when light waves encounter obstacles or openings, they bend around the edges. Circular apertures create a specific pattern because of their symmetrical shape.

In our exercise, we are observing a phenomenon called diffraction. Light passing through the circular aperture spreads out, forming a series of concentric rings on the screen. These are the diffraction patterns that are often seen in experiments. The size of these rings depends on factors such as the wavelength of the light and the size of the aperture. The main observation is that smaller apertures create larger diffraction patterns.

Monochromatic Light

Monochromatic light is light that is of one single wavelength or color. Examples are the specific colors you might observe in lasers, such as red or green. In this exercise, the wavelength used is 490 nm, which falls in the blue spectrum.

Using monochromatic light is crucial for creating clear diffraction patterns. If you were to use white light, which contains multiple wavelengths, the resulting pattern would be complex and not as easy to analyze. With monochromatic light, the diffraction patterns are sharp, making it easier to measure distances between dark and light regions on the screen.

These measurements help in determining the properties of the light's journey through the circular aperture, such as confirming the diameter of the aperture itself. Monochromatic light is chosen for its simplicity and the clarity it brings to experiments and observations.

Angular Position of Minima

The angular position of minima in a diffraction pattern is an important concept when studying wave behavior. Minima are the dark rings that appear on the screen, where no light reaches. This occurs because of destructive interference, where light waves cancel each other out.

The formula for the angular position of the minima in our circular aperture scenario is given by \( \theta = 1.22\frac{\lambda}{D} \), where \( \theta \) is the angular position, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture.

The calculation for the angular position allows us to understand how far apart these dark rings, or minima, are. In practical terms, it aids in calculating the diameter of the aperture. As shown in the original exercise, the distance between these minima is needed to solve for the aperture size. The clearer we define these positions, the more accurate our calculations will be.