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Chapter 3: Problem 89

Determine what happens to the double-slit interference pattern if one of the slits is covered with a thin, transparent film whose thickness is \(\lambda /[2(n-1)]\), where \(\lambda\) is the wavelength of the incident light and \(n\) is the index of refraction of the film.

Short Answer

Expert verified
When one of the slits in a double-slit interference experiment is covered with a thin transparent film of thickness \(\frac{\lambda}{2(n-1)}\), the path length of the light wave passing through that slit is increased by \(\frac{\lambda}{n-1}\), leading to a new path difference of \(d \sin \theta + \frac{\lambda}{n-1}\). This will cause a shift in the interference pattern and may change the positions of the bright and dark fringes, depending on the values of \(d\), \(n\), and \(\theta\).

Step by step solution

01

Understand the double-slit interference pattern

A double-slit interference pattern occurs when light waves pass through two slits and interfere with each other. The interference pattern is a series of bright fringes (constructive interference) and dark fringes (destructive interference) due to the differences in the path lengths of the light waves traveling from each slit to a point on the screen.
02

Determine the path difference between the light waves from the two slits

Let's consider the situation where the light waves travel from each slit and meet at a point on the screen. The path difference between the light waves from the two slits determines whether the interference will be constructive (bright fringe) or destructive (dark fringe). The path difference is given by: Path difference = \(d \sin \theta\), where \(d\) is the distance between the slits and \(\theta\) is the angle between the central maximum and the fringe on the screen.
03

Calculate the phase difference due to the thin film

When one of the slits is covered with a thin transparent film, the path length of the light wave passing through that slit is increased. The additional path length is given by: Additional path length = 2 * (thickness of the film) Since the thickness of the film is given by \(\lambda /[2(n-1)]\), the additional path length can be written as: Additional path length = 2 * \(\frac{\lambda}{2(n-1)}\) = \(\frac{\lambda}{n-1}\)
04

Determine the new path difference with the thin film

With the thin film covering one of the slits, the new path difference between the light waves from the two slits becomes: New path difference = \(d \sin \theta + \frac{\lambda}{n-1}\)
05

Analyze the effect of the thin film on the interference pattern

To determine the effect of the thin film on the interference pattern, we need to examine the conditions for constructive and destructive interference: Constructive interference (bright fringe) occurs when: New path difference = \(m \lambda\), where \(m\) is an integer. Destructive interference (dark fringe) occurs when: New path difference = \((m + 1/2) \lambda\), where \(m\) is an integer. So, depending on the values of \(d\), \(n\), and \(\theta\), the interference pattern may shift, and the positions of the bright and dark fringes may change. The specific change in the pattern will depend on the parameters of the problem.

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Most popular questions from this chapter

After a minor oil spill, a think film of oil \((n=1.40)\) of thickness 450 nm floats on the water surface in a bay. (a) What predominant color is seen by a bird flying overhead? (b) What predominant color is seen by a seal swimming underwater?

Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by \(0.100 \mathrm{mm}\).

A Michelson interferometer with a He-Ne laser light source \((\lambda=632.8 \mathrm{nm})\) projects its interference pattern on a screen. If the movable mirror is caused to move by \(8.54 \mu \mathrm{m},\) how many fringes will be observed shifting through a reference point on a screen?

What is the angular width of the central fringe of the interference pattern of (a) 20 slits separated by \(d=2.0 \times 10^{-3} \mathrm{mm} ?\) (b) 50 slits with the same separation? Assume that \(\lambda=600 \mathrm{nm}\).

In a thermally stabilized lab, a Michelson interferometer is used to monitor the temperature to ensure it stays constant. The movable mirror is mounted on the end of a 1.00 -m-long aluminum rod, held fixed at the other end. The light source is a He Ne laser, \(\lambda=632.8 \mathrm{nm}\). The resolution of this apparatus corresponds to the temperature difference when a change of just one fringe is observed. What is this temperature difference?
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