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Young's experiment is performed with light from excited helium atoms (\(\lambda\) = 502 nm). Fringes are measured carefully on a screen 1.20 m away from the double slit, and the center of the 20th fringe (not counting the central bright fringe) is found to be 10.6 mm from the center of the central bright fringe. What is the separation of the two slits?

Short Answer

Expert verified
The separation of the slits is approximately 0.114 mm.

Step by step solution

01

Understand the formula

In Young's double-slit experiment, the equation for fringe position is \( y_m = \frac{m \cdot \lambda \cdot D}{d} \) where \( y_m \) is the distance from the central maximum to the m-th order fringe, \( m \) is the order of the fringe, \( \lambda \) is the wavelength of the light, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.
02

Identify given values

We are given:- \( \lambda = 502 \) nm (convert to meters: \( 502 \times 10^{-9} \) m),- Distance to screen \( D = 1.20 \) m,- Distance of the 20th fringe \( y_{20} = 10.6 \) mm (convert to meters: \( 10.6 \times 10^{-3} \) m),- Order of the fringe \( m = 20 \).
03

Rearrange the formula to find slit separation

To find the slit separation \( d \), rearrange the formula to \( d = \frac{m \cdot \lambda \cdot D}{y_m} \). This will allow us to solve for \( d \) using the values identified in Step 2.
04

Substitute in the known values

Substitute \( m = 20 \), \( \lambda = 502 \times 10^{-9} \) m, \( D = 1.20 \) m, and \( y_m = 10.6 \times 10^{-3} \) m into the rearranged formula: \[ d = \frac{20 \cdot 502 \times 10^{-9} \cdot 1.20}{10.6 \times 10^{-3}} \].
05

Calculate the slit separation

Perform the calculation: \[ d = \frac{20 \cdot 502 \times 10^{-9} \cdot 1.20}{10.6 \times 10^{-3}} = \frac{12048 \times 10^{-9}}{10.6 \times 10^{-3}} \approx 1.1373 \times 10^{-4} \text{ m} \] or \( 0.11373 \text{ mm} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fringe Separation Calculation
To calculate the fringe separation in Young's Double-Slit Experiment, we can use the formula \( y_m = \frac{m \cdot \lambda \cdot D}{d} \). Let's break down what each part means. Here, \( y_m \) is the distance from the central maximum to the m-th order fringe, \( m \) stands for the order of the fringe (e.g., 20th fringe), \( \lambda \) is the wavelength of the light used in the experiment, \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits.

By rearranging this formula, you can solve for the slit separation \( d \) by using known values of the fringe position, order, wavelength, and screen distance:
  • First, identify the order \( m \) and substitute it into the rearranged formula \( d = \frac{m \cdot \lambda \cdot D}{y_m} \).
  • Next, plug in the values for \( \lambda \), \( D \), and \( y_m \) accordingly.
  • Finally, calculate \( d \) to find how far the two slits are apart, allowing us to understand the pattern of the light fringes formed on the screen.
This method reveals the fascinating interplay between light properties and physical separations, providing insights into wave-interference behaviors.
Wavelength of Light
The wavelength of light, denoted by \( \lambda \), is a crucial part of any interference experiment, like Young's Double-Slit. It determines the color of the light and significantly affects the interference pattern formed. In the original exercise, excited helium atoms emit light with a wavelength of 502 nm (nanometers), which we must convert to meters to use in our calculations, giving us \( 502 \times 10^{-9} \) meters.

Wavelength is the distance between consecutive peaks (or troughs) of a wave. Visible light, like the helium light in the experiment, is made up of different wavelengths within a range. These differences are what give light its color:
  • Shorter wavelengths (e.g., blue light) have higher energy.
  • Longer wavelengths (e.g., red light) have lower energy.
  • The wavelength directly affects how the light interferes with itself when it passes through slits.
Understanding the wavelength of light used ensures we're accurately depicting wave behaviors in double-slit setups, leading to precise interference patterns.
Interference Pattern
An interference pattern is a series of light and dark bands or fringes, resulting from constructive and destructive interference of light waves. In Young's Double-Slit Experiment, light waves pass through two slits, creating a pattern on a screen due to this wave interference.

Here's how it works:
  • Constructive Interference: Occurs when waves from both slits arrive in phase, resulting in bright fringes.
  • Destructive Interference: Occurs when waves from both slits arrive out of phase, resulting in dark fringes.
These alternating bright and dark fringes form the signature interference pattern, measurable as fringe separation. By measuring this pattern, we gain insights into wave properties of light and confirm its wave-like behavior. This phenomenon beautifully illustrates the principles of wave interference, a hallmark in the study of optics.

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Most popular questions from this chapter

Two slits spaced 0.260 mm apart are 0.900 m from a screen and illuminated by coherent light of wavelength 660 nm. The intensity at the center of the central maximum (\(\theta\) = 0\(^\circ\)) is \(I_0\) . What is the distance on the screen from the center of the central maximum (a) to the first minimum; (b) to the point where the intensity has fallen to \(I_0\) /2?

A plastic film with index of refraction 1.70 is applied to the surface of a car window to increase the reflectivity and thus to keep the car's interior cooler. The window glass has index of refraction 1.52. (a) What minimum thickness is required if light of wavelength 550 nm in air reflected from the two sides of the film is to interfere constructively? (b) Coatings as thin as that calculated in part (a) are difficult to manufacture and install. What is the next greater thickness for which constructive interference will also occur?

In your research lab, a very thin, flat piece of glass with refractive index 1.40 and uniform thickness covers the opening of a chamber that holds a gas sample. The refractive indexes of the gases on either side of the glass are very close to unity. To determine the thickness of the glass, you shine coherent light of wavelength \(\lambda_0\) in vacuum at normal incidence onto the surface of the glass. When \(\lambda_0\) = 496 nm, constructive interference occurs for light that is reflected at the two surfaces of the glass. You find that the next shorter wavelength in vacuum for which there is constructive interference is 386 nm. (a) Use these measurements to calculate the thickness of the glass. (b) What is the longest wavelength in vacuum for which there is constructive interference for the reflected light?

Coherent light of frequency \(6.32 \times10^{14}\) Hz passes through two thin slits and falls on a screen 85.0 cm away. You observe that the third bright fringe occurs at \(\pm\)3.11 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?

A radio transmitting station operating at a frequency of 120 MHz has two identical antennas that radiate in phase. Antenna B is 9.00 m to the right of antenna A. Consider point P between the antennas and along the line connecting them, a horizontal distance x to the right of antenna A. For what values of x will constructive interference occur at point P?

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