Question

Two slits spaced 0.450 mm apart are placed 75.0 cm from a screen. What is the distance between the second and third dark lines of the interference pattern on the screen when the slits are illuminated with coherent light with a wavelength of 500 nm?

Short answer

The distance is approximately 0.083 mm.

Step-by-step solution

Understand the Concepts

Interference patterns involve alternating light and dark bands. Dark bands (minima) occur at positions where the path difference between light from the two slits is a half-integer multiple of the wavelength, i.e., \[ d \sin \theta = (m + \frac{1}{2})\lambda \]where \(d\) is the slit separation, \(\theta\) is the angle of the dark fringe, \(m\) is the order of the minimum (1, 2 for second and third dark lines), and \(\lambda\) is the wavelength of the light.

Calculate the Angle for the Second Dark Line

For the second dark line, \(m = 1\). Using the formula for the dark fringes, \[ 0.450 \times 10^{-3} \sin \theta_2 = (1 + \frac{1}{2})(500 \times 10^{-9}) \]Solve for \(\theta_2\):\[ \sin \theta_2 = \frac{1.5 \times 500 \times 10^{-9}}{0.450 \times 10^{-3}} \approx 1.67 \times 10^{-3} \]Thus, \(\theta_2 \approx \sin^{-1}(1.67 \times 10^{-3})\).

Calculate the Angle for the Third Dark Line

For the third dark line, \(m = 2\). Use the dark fringe pattern formula:\[ 0.450 \times 10^{-3} \sin \theta_3 = (2 + \frac{1}{2})(500 \times 10^{-9}) \]Solve for \(\theta_3\):\[ \sin \theta_3 = \frac{2.5 \times 500 \times 10^{-9}}{0.450 \times 10^{-3}} \approx 2.78 \times 10^{-3} \]Thus, \(\theta_3 \approx \sin^{-1}(2.78 \times 10^{-3})\).

Find the Positions on the Screen

Use the small angle approximation, where \(\sin \theta \approx \tan \theta \approx \theta\) and the formula for distance from the center, \[ y = L \tan \theta \approx L \theta \]For \(\theta_2\) and \(\theta_3\) calculated earlier, with \(L = 75.0 \times 10^{-2} \) m:\[ y_2 = 75.0 \times 10^{-2} \cdot 1.67 \times 10^{-3} \approx 1.2525 \times 10^{-1} \text{ mm} \]\[ y_3 = 75.0 \times 10^{-2} \cdot 2.78 \times 10^{-3} \approx 2.085 \times 10^{-1} \text{ mm} \]

Determine the Distance Between Second and Third Dark Lines

Subtract the positions of the second and third dark lines:\[ \Delta y = y_3 - y_2 = 0.2085 - 0.12525 \approx 0.08325 \text{ mm} \]

Conclusion

The distance between the second and third dark lines on the screen is approximately 0.083 mm.

Key concepts

Young's Double-Slit Experiment

Young's Double-Slit Experiment is a classic physics experiment demonstrating the wave nature of light. When light passes through two closely spaced slits, it creates an interference pattern on a screen placed behind the slits. This pattern consists of alternating bright and dark areas called fringes.
If the light coming through the slits was considered as particles, you would expect two distinct lines on the screen because particles travel straight. However, by observing a pattern of multiple lines, it suggests that light behaves like a wave, experiencing constructive and destructive interference.
Understanding this collaboration of light waves through the slits remains a cornerstone of optical and quantum physics.

Optical Path Difference

The Optical Path Difference (OPD) is crucial in analyzing interference patterns. It refers to the difference in the travel paths of light waves coming from the two slits to a common point on the screen.
As waves travel different distances, the OPD leads to either amplification or cancellation of the waves. This interference is mathematically expressed as
\[ d \sin \theta = m\lambda \quad \text{for bright fringes} \]
and
\[ d \sin \theta = \left(m + \frac{1}{2}\right)\lambda \quad \text{for dark fringes} \]
Here, \(d\) is the slit separation, \(\lambda\) is the wavelength of the light, and \(m\) is the fringe order.
The OPD concept helps explain why you see specific patterns of light and dark bands.

Wavelength and Slit Separation

Understanding the relationship between wavelength and slit separation is vital in Young's Double-Slit Experiment. The wavelength \(\lambda\) is the distance over which the wave's shape repeats. In this experiment, knowing the wavelength of the incoming light is essential to calculate where the interference fringes will appear.
Slit separation \(d\) is the physical distance between the two slits. This separation plays a critical role in determining the spacing of the interference pattern. As the slit separation increases, the fringes move closer together, and vice versa.
The slit separation and wavelength together shape the entire interference pattern according to the formula:
\[ d \sin \theta = \left(m + \frac{1}{2}\right)\lambda \]
where \(\theta\) is the angle at which a fringe occurs.

Dark Fringe Calculation

Dark fringes, which are the areas of minima in an interference pattern, occur where the path difference leads to destructive interference. To calculate these dark fringes, the formula \[ d \sin \theta = \left(m + \frac{1}{2}\right)\lambda \] is used.
Here, \(m\) denotes the order of the dark fringe being calculated. For example, \(m = 1\) for the second dark line and \(m = 2\) for the third, and so forth.
Once the angle \(\theta\) for these dark fringes is found using the above formula, you can convert it to a linear distance \(y\) on the screen using \[ y = L \tan \theta \approx L \theta \].
This conversion is possible due to the small angle approximation where \(\sin \theta \approx \theta\), allowing you to find the position where the light waves cancel out, leading to the dark fringes.