Question

Two thin parallel slits that are 0.0116 mm apart are illuminated by a laser beam of wavelength 585 nm. (a) On a very large distant screen, what is the \(total\) number of bright fringes (those indicating complete constructive interference), including the central fringe and those on both sides of it? Solve this problem without calculating all the angles! (\(Hint\): What is the largest that sin u can be? What does this tell you is the largest value of \(m\)?) (b) At what angle, relative to the original direction of the beam, will the fringe that is most distant from the central bright fringe occur?

Short answer

There are 39 bright fringes. The angle for the most distant fringe is approximately 73.5°.

Step-by-step solution

Understand the Concept

This problem is about the double-slit experiment, which demonstrates interference patterns. The formula for bright fringes is given by \( d \sin \theta = m\lambda \), where \( d \) is the distance between the slits, \( \theta \) is the angle of the fringe, \( m \) is the fringe order (an integer), and \( \lambda \) is the wavelength of the light.

Determine the Maximum Value of \( m \)

Since \( \sin \theta \) cannot exceed 1, the maximum value of \( m \) for which a fringe can occur is when \( m\lambda \leq d \). Thus, the maximum \( m \) is determined as \( m = \frac{d}{\lambda} \). Substituting given values: \( d = 0.0116 \) mm = \( 11.6 \times 10^{-6} \) m and \( \lambda = 585 \) nm = \( 585 \times 10^{-9} \) m, which calculates to \( m = \frac{11.6 \times 10^{-6}}{585 \times 10^{-9}} \approx 19.82 \). This means \( m \) can be at most 19 each side.

Calculate Total Number of Bright Fringes

For \( m = 0 \), we have the central bright fringe. For \( m = -19 \) to \( m = 19 \), we include fringes on both sides. Therefore, the number of bright fringes is \( 2 \times 19 + 1 = 39 \).

Determine the Angle for the Most Distant Fringe

For the most distant fringe (\( m = 19 \)), we use \( \sin \theta = \frac{m\lambda}{d} \). Substituting \( m = 19 \), \( \lambda = 585 \times 10^{-9} \) m, and \( d = 11.6 \times 10^{-6} \) m into the formula gives \( \sin \theta \approx \frac{19 \times 585 \times 10^{-9}}{11.6 \times 10^{-6}} \approx 0.957 \). Thus, \( \theta = \arcsin(0.957) \approx 73.5^\circ \).

Key concepts

Constructive Interference

In the double-slit experiment, constructive interference happens when waves from two slits meet in phase. This means that the crests and troughs of the waves align perfectly, amplifying the light and creating a bright spot on the screen. This bright spot is known as a bright fringe. Constructive interference occurs at specific angles, where the path difference between waves from each slit equals an integer multiple of the wavelength.
In our problem, the condition for constructive interference is given by the formula: \[ d \sin \theta = m\lambda \] Here, \(d\) is the distance between the slits, \(\theta\) is the angle at which the constructive interference occurs, \(m\) is the order of the bright fringe, and \(\lambda\) is the wavelength of the light. The order \(m\) can be positive, negative, or zero, corresponding to different bright fringes on the screen.
Understanding this concept is key to predicting where and how light intensifies in patterns.

Interference Patterns

Interference patterns are a series of light and dark bands on a screen resulting from the superposition of light waves. The double-slit experiment creates these patterns when coherent light, like from a laser, passes through two closely spaced slits. On the screen, you'll notice alternating bright and dark stripes. Bright stripes occur where waves constructively interfere, while dark stripes are regions of destructive interference, where waves cancel each other out.
The spacing and intensity of these patterns depend on factors such as:

• Distance between the slits
• Wavelength of the light used
• Distance from the slits to the screen

By calculating the angles for various orders of interference, our problem demonstrates how to predict the positions of bright fringes in the interference pattern.

Wavelength of Light

The wavelength of light, denoted by \(\lambda\), is a crucial factor in determining interference patterns in the double-slit experiment. It is the distance between successive peaks of a wave. In our exercise, \(\lambda\) is given as 585 nm, where a nanometer (nm) is one-billionth of a meter, a common unit for measuring wavelengths of light.
Shorter wavelengths, like blue light, require more precise slits to create extensive interference patterns. Longer wavelengths, like red light, typically result in more spread-out patterns. The light's wavelength determines how waves interfere with each other and effectively dictates the position and number of bright and dark fringes observed.
Understanding how wavelength affects interference helps explain why the formula \(d \sin \theta = m\lambda\) is vital in calculating where interference maxima will appear.

Angle of Diffraction

In the context of interference patterns, the angle of diffraction \(\theta\) relates directly to where bright fringes appear on the projection screen. The angle \(\theta\) is measured from the normal—or perpendicular line—to the screen or central fringe. Depending on the conditions and parameters of the experiment, such as light wavelength and distance between slits, this angle dictates the exact positioning of bright spots.The relationship is defined by the equation \[d \sin \theta = m\lambda\] which allows the calculation of \(\theta\) at different fringe orders \(m\). For example, in our problem, by plugging known values into this equation, we determine that the most distant fringe creates an angle of about 73.5° relative to the original light direction.Understanding how to derive this angle without complex calculations is essential, as it helps visualize and predict how light behaves as it encounters obstacles.