Problem 49 - University Physics with Modern Physics

Question

Red light with wavelength 700 nm is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe (\(m\) = 32) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

Step-by-Step Solution

Answer

Possible wavelengths are integer multiples of \\(22400\\, ext{nm}/(m'+0.5)\\) within visible range. Slit spacing is not needed.

Step 1: Understanding the Problem

A red light with wavelength 700 nm creates an interference pattern. For the third bright fringe (\(m = 32\)) of the red pattern to appear pure red, the other wavelength must have a dark fringe at this position. This requires satisfying the condition for destructive interference for the other light.

Step 2: Defining the Condition for Bright Fringes

For bright fringes in a double-slit experiment, the path difference must be an integer multiple of the wavelength, \(d \sin \theta = m \lambda\), where \(d\) is the slit spacing, \(\theta\) is the angle of the fringe, and \(m\) is the order of the fringe.

Step 3: Defining the Condition for Dark Fringes

For dark fringes, the condition is \(d \sin \theta = (m' + 0.5) \lambda'\), where \(m'\) is an integer representing the order of the dark fringe and \(\lambda'\) is the wavelength of the second light. For a pure red at \(m = 32\), we need a dark fringe of the other light at the same position.

Step 4: Finding the Required Relationship

Since both the red light and other light are passing through the same apparatus, the angle \(\theta\) is the same: \(32 \lambda_{red} = (m'+0.5) \lambda_{other}\). Thus, \((m'+0.5) \lambda_{other} = 32 \times 700 \, \text{nm}\).

Step 5: Solving for Possible Wavelengths

Isolating for \(\lambda_{other}\), we get \(\lambda_{other} = \frac{32 \times 700}{m' + 0.5}\). As the final expression shows, possible wavelengths for \(\lambda_{other}\) need to be visible light (approximately 380 - 750 nm). This limits the value of \(m'\).

Step 6: Determining if Slit Spacing is Required

The question asks if slit spacing is required. Since both wavelengths are calculated based on path difference, which relies only on their order terms and proportionate wavelengths, the actual slit spacing is not needed. It cancels out when determining the ratio of the distances involved.

Key Concepts

Interference Pattern

When light passes through two closely spaced slits, it creates an **interference pattern** on a screen behind the slits. This pattern is made of alternating bands of bright and dark areas or fringes. The bright fringes occur where the light waves from the slits reinforce each other, known as constructive interference, while dark fringes occur where the waves cancel each other out, known as destructive interference.

The interference pattern is a fundamental aspect of the double-slit experiment.

• Bright fringes are observed at points where the path difference between the slits equals an integer multiple of the wavelength: \(d \sin \theta = m \lambda\), where \(d\) represents the slit spacing, \(\theta\) is the angle to the fringe, and \(m\) is the fringe order.
• Dark fringes occur when the path difference is a half-integer multiple of the wavelength: \(d \sin \theta = (m' + 0.5) \lambda'\).

The exact positions of these bright and dark fringes are determined by the geometry of the setup and the light's wavelength.

Wavelength of Light

The **wavelength of light** is a crucial parameter in determining the nature of the interference pattern. It is the distance between successive peaks of a wave and determines the color of the light.

In the context of the double-slit experiment, the wavelength affects both the spacing and intensity of the interference fringes. Different wavelengths produce differently spaced patterns because they influence the path difference equation:

• For example, red light with a longer wavelength (700 nm) creates a more widely spaced pattern compared to blue light with a shorter wavelength.
• The interference equation for bright fringes is \(d \sin \theta = m \lambda\), while for dark fringes, it's \(d \sin \theta = (m' + 0.5) \lambda'\).

The problem you've explored involved determining the wavelength of another visible light resulting in a dark fringe at a particular position aligning with a pure red fringe produced by light of 700 nm.

Destructive Interference

**Destructive interference** is a key factor in producing the dark fringes we see in an interference pattern. It occurs when two waves meet in such a way that their crests coincide with the troughs of the other.

In mathematical terms, this happens when the path difference between the waves is a half-odd multiple of their wavelength, causing them to cancel each other out. This condition is expressed by the formula:

• Dark fringe condition: \(d \sin \theta = (m' + 0.5) \lambda'\), where \(m'\) is the order of the fringe.

In your exercise, it illustrates how the center of a third bright red fringe appears to be pure red because other wavelengths undergo destructive interference at this point, creating a position where no light from the secondary source is observed. This aspect of interference patterns allows for the clear differentiation of specific wavelengths at particular points on a screen.