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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?

When viewing a piece of art that is behind glass, one often is affected by the light that is reflected off the front of the glass (called \(glare\)), which can make it difficult to see the art clearly. One solution is to coat the outer surface of the glass with a film to cancel part of the glare. (a) If the glass has a refractive index of 1.62 and you use Ti\(O^2\) , which has an index of refraction of 2.62, as the coating, what is the minimum film thickness that will cancel light of wavelength 505 nm? (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed. The plates are illuminated at normal incidence by 546-nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge

What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 nm? The index of refraction of the film is 1.33, and there is air on both sides of the film.

After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of 1.45. After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 nm has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

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