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Chapter 27: Problem 9

A sheet that is made of plastic \((n=1.60)\) covers one slit of a double slit (see the drawing). When the double slit is illuminated by monochromatic light \(\left(\lambda_{\text {vacuum }}=586 \mathrm{nm}\right),\) the center of the screen appears dark rather than bright. What is the minimum thickness of the plastic?

Short Answer

Expert verified
The minimum thickness of the plastic is 488.33 nm.

Step by step solution

01

Understand the problem situation

We're given a double-slit experiment where a plastic sheet with refractive index \(n = 1.60\) covers one of the slits. Monochromatic light of wavelength \( \lambda = 586 \text{ nm} \) is used, and the center of the screen appears dark, indicating destructive interference. We need to determine the minimum thickness of the plastic sheet so that the path difference leads to this result.
02

Conceptualize destructive interference

For the center of the screen to appear dark, there must be a phase difference equivalent to half a wavelength (a path difference of \(m + 0.5\) wavelengths, where \(m\) is an integer). In this case, the path difference is due to the optical thickness (\(t'(n-1)\)) of the plastic sheet.
03

Calculate the optical path difference

The optical path difference needed for destructive interference is \((m + 0.5) \cdot \lambda\). For minimum thickness, let's consider \(m = 0\). Therefore, the optical path difference is \(0.5\cdot\lambda\). In terms of thickness and refractive index, this equation becomes \(t(n - 1) = 0.5 \cdot \lambda\).
04

Solve for thickness \(t\)

Rearrange the equation from Step 3: \[ t(n - 1) = 0.5 \cdot \lambda \] Where \(\lambda\) is the wavelength in a vacuum. Substituting \(n = 1.60\) and \( \lambda = 586 \text{ nm} \): \[ t = \frac{0.5 \cdot 586 \text{ nm}}{1.60 - 1} = \frac{293.0 \text{ nm}}{0.60} \] Calculate: \[ t = 488.33 \text{ nm} \] This is the minimum thickness that will result in a dark center on the screen.

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Key Concepts

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

Understanding Optical Path Difference
In a double-slit experiment, light waves travel different paths to reach a point on a screen. The difference in these paths is called the "optical path difference." This difference determines whether the light waves will interfere constructively or destructively.
An optical path difference is calculated by considering both the geometric path and the material through which the light passes. The formula can be expressed as:\[ \text{Optical Path Difference} = t(n-1) \]
Here, \( t \) is the material's thickness, and \( n \) is the refractive index. By controlling this difference, we can manipulate the interference pattern on a screen. To achieve destructive interference at the screen's center, the optical path difference must be half a wavelength. This calculation helps find the required thickness of the plastic covering one of the slits.
Exploring Destructive Interference
Destructive interference occurs when the crest of one wave meets the trough of another, canceling each other out. This results in a dark spot on the screen. For the experiment's central point to appear dark, the waves from both slits need to be out of phase by half a wavelength.
  • When two waves are half a wavelength out of phase, they interfere destructively.
  • This means the optical path difference is \( \frac{1}{2} \lambda \), where \( \lambda \) is the wavelength.
Destructive interference in a double-slit setup provides valuable insight into wave behavior. By knowing this concept, you can determine how to manipulate the interference pattern by introducing elements like a plastic sheet with a specific thickness.
Role of Refractive Index
The refractive index \( n \) indicates how much light slows down when passing through a material. This factor significantly influences optical path differences.
  • A higher refractive index means light travels slower through the material.
  • This also increases the effective optical path, contributing to interference effects.
Given a refractive index of 1.60, the plastic sheet causes light to slow down more compared to air, affecting the phase and resulting in destructive interference. Calculating the impact of refractive index helps in designing experiments and understanding how materials interact with light, ultimately leading to precise control over interference patterns.

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Most popular questions from this chapter

Interactive LearningWare 27.2 at provides some pertinent background for this problem. A transparent film \((n=1.43)\) is deposited on a glass plate \((n=1.52)\) to form a nonreflecting coating. The film has a thickness that is \(1.07 \times 10^{-7} \mathrm{~m}\). What is the longest possible wavelength (in vacuum) of light for which this film has been designed?

A uniform layer of water \((n=1.33)\) lies on a glass plate \((n=1.52)\). Light shines perpendicularly on the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the light is \(432 \mathrm{~nm}\) in vacuum and also when it is \(648 \mathrm{~nm}\) in vacuum. (a) Obtain the minimum thickness of the film. (b) Assuming that the film has the minimum thickness and that the visible spectrum extends from 380 to \(750 \mathrm{~nm}\), determine the visible wavelength(s) (in vacuum) for which the film appears completely dark.

A hunter who is a bit of a braggart claims that, from a distance of \(1.6 \mathrm{~km}\), he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{~nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{~mm}\), the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of \(498 \mathrm{~nm}\).

Two gratings \(A\) and \(B\) have slit separations \(d_{A}\) and \(d_{B}\), respectively. They are used with the same light and the same observation screen. When grating A is replaced with grating \(\mathrm{B}\), it is observed that the first-order maximum of \(\mathrm{A}\) is exactly replaced by the secondorder maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating \(\mathrm{A}\) and the principal maxima of B that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

A spotlight sends red light (wavelength \(=694.3 \mathrm{nm}\) ) to the moon. At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{~m}\) away, the light strikes a reflector left there by astronauts. The reflected light returns to the earth, where it is detected. When it leaves the spotlight, the circular beam of light has a diameter of about \(0.20 \mathrm{~m}\), and diffraction causes the beam to spread as the light travels to the moon. In effect, the first circular dark fringe in the diffraction pattern defines the size of the central bright spot on the moon. Determine the diameter (not the radius) of the central bright spot on the moon.
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