Problem 18 - University Physics with Modern Physics

Question

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{Hz}\) passes through two thin slits and falls on a screen 85.0 \(\mathrm{cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

Step-by-Step Solution

Answer

(a) The slits are 39.0 micrometers apart. (b) The third dark fringe occurs at ±3.63 cm.

Step 1: Identify Given Parameters

We start by listing the given parameters:- The frequency of light, \( f = 6.32 \times 10^{14} \mathrm{Hz} \).- The distance from the slits to the screen, \( L = 85.0 \mathrm{cm} = 0.85 \mathrm{m} \).- Position of the third bright fringe, \( y_3 = 3.11 \mathrm{cm} = 0.0311 \mathrm{m} \).

Step 2: Calculate the Wavelength of Light

The wavelength of light (\( \lambda \)) is calculated using the speed of light \( c = 3.00 \times 10^8 \mathrm{m/s} \) and the formula \( \lambda = \frac{c}{f} \):\[\lambda = \frac{3.00 \times 10^8}{6.32 \times 10^{14}} = 4.75 \times 10^{-7} \mathrm{m}\]

Step 3: Apply the Formula for Bright Fringes

The condition for the nth bright fringe in a double-slit experiment is given by:\[y_n = \frac{n \cdot \lambda \cdot L}{d}\]Here, \( n = 3 \), because it's the third bright fringe. We now solve for slit separation \( d \):\[d = \frac{n \cdot \lambda \cdot L}{y_3} = \frac{3 \cdot 4.75 \times 10^{-7} \cdot 0.85}{0.0311}\]

Step 4: Calculate the Slit Separation

Substitute the known values into the formula from Step 3:\[d = \frac{3 \cdot 4.75 \times 10^{-7} \cdot 0.85}{0.0311} = 3.90 \times 10^{-5} \mathrm{m}\]Thus, the slit separation \( d \) is 39.0 micrometers.

Step 5: Calculate the Position of the Third Dark Fringe

The position for dark fringes is given by:\[y_m = \frac{(m + 0.5) \cdot \lambda \cdot L}{d}\]For the third dark fringe, \( m = 3 \). Substitute \( d = 3.90 \times 10^{-5} \mathrm{m} \):\[y_m = \frac{(3 + 0.5) \cdot 4.75 \times 10^{-7} \cdot 0.85}{3.90 \times 10^{-5}}\]

Step 6: Calculate Distance for Third Dark Fringe

Substitute the values into the formula used in Step 5:\[y_m = \frac{3.5 \cdot 4.75 \times 10^{-7} \cdot 0.85}{3.90 \times 10^{-5}} = 3.63 \mathrm{cm}\]The third dark fringe occurs at \( \pm 3.63 \mathrm{cm} \) from the central fringe.

Key Concepts

Coherent Light

In the context of the double-slit experiment, coherent light refers to light waves that maintain a constant phase relationship. This coherence is essential for creating a stable pattern of interference on a screen, comprising both bright and dark fringes.
Coherent light typically originates from lasers or specially designed setups, where light waves vibrate in synchronization.
By maintaining this uniform phase condition, coherent light allows for the clear observation of interference patterns, an important aspect of the double-slit experiment.
This phase consistency results in reinforcement (constructive interference) and cancellation (destructive interference), giving rise to the observable fringes.

Bright and Dark Fringes

In a double-slit experiment, bright fringes, also known as maxima, occur at points where the light waves from the two slits arrive in phase, undergoing constructive interference. Bright fringes appear as alternating bands of light on the screen. These occur at specific points calculated using the formula: \[y_n = \frac{n \cdot \lambda \cdot L}{d}\] where represents the order of the fringe, \(\lambda\) is the wavelength, \(L\) is the distance to the screen, and \(d\) is the slit separation.
Conversely, dark fringes, or minima, occur at destructive interference points where light waves arrive out of phase, canceling each other out. The position of dark fringes is given by: \[y_m = \frac{(m + 0.5) \cdot \lambda \cdot L}{d}\] where is the fringe order, and \((m + 0.5)\) accounts for the half-wavelength offset typical in destructive interference.
Understanding these patterns allows us to explore many properties of light, including its wave nature.

Wavelength Calculation

Wavelength calculation is a key step in analyzing interference patterns from the double-slit experiment. The wavelength of light is directly related to its frequency and can be calculated using the speed of light \(c = 3.00 \times 10^8 \text{m/s}\).
The formula \( \lambda = \frac{c}{f} \) is used, where \( f \) denotes the frequency. In this exercise, for a frequency of \( 6.32 \times 10^{14} \text{Hz} \), the wavelength can be calculated as follows:
\[\lambda = \frac{3.00 \times 10^8}{6.32 \times 10^{14}} = 4.75 \times 10^{-7} \text{m}\]
This calculation is crucial because the wavelength is pivotal in determining bright and dark fringe positions, thus enabling us to determine physical characteristics like slit separation.

Slit Separation

The separation of the two slits in a double-slit experiment is pivotal for interpreting interference patterns. It directly influences the spacing and position of bright and dark fringes. Slit separation \(d\) can be derived from the formula for bright fringes:
\[d = \frac{n \cdot \lambda \cdot L}{y_n}\]
For a given fringe order \(n\), the wavelength \(\lambda\), and known distances \(L\) and \(y_n\), \(d\) provides the required separation. In the provided exercise, it is calculated as \(3.90 \times 10^{-5} \text{m}\) or 39.0 micrometers, illustrating the small scale and precision necessary for the double-slit experiment.
This concept not only deepens our understanding of wave interference but also highlights the precise nature of optical instrumentation and setups.