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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 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?

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50. At room temperature (20.0\(^\circ\)C), this film is just thick enough for light with wavelength 582.4 nm reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to 170\(^\circ\)C, you find that the film cancels reflected light with wavelength 588.5 nm. What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

Two small stereo speakers \(A\) and \(B\) that are 1.40 m apart are sending out sound of wavelength 34 cm in all directions and all in phase. A person at point \(P\) starts out equidistant from both speakers and walks so that he is always 1.50 m from speaker \(B\) (Fig. E35.1). For what values of x will the sound this person hears be (a) maximally reinforced, (b) cancelled? Limit your solution to the cases where x \(\leq\) 1.50 m.

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at \(\pm\)19.0\(^\circ\) with the original direction of the beam, as viewed on a screen far from the slits. (a) What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits? (b) What is the smallest angle, relative to the original direction of the laser beam, at which the intensity of the light is \(1 \over 10\) the maximum intensity on the screen?

Suppose you illuminate two thin slits by monochromatic coherent light in air and find that they produce their first interference \(minima\) at \(\pm35.20^\circ\) on either side of the central bright spot. You then immerse these slits in a transparent liquid and illuminate them with the same light. Now you find that the first minima occur at \(\pm19.46^\circ\) instead. What is the index of refraction of this liquid?

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