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Chapter 35: Problem 49

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 \(\frac{1}{10}\) the maximum intensity on the screen?

Short Answer

Expert verified
The ratio of the slit separation to the wavelength comes out to be approximately 0.325 and the smallest angle relative to the original direction is obtained by solving the second equation, which will vary specifically with the parameters given.

Step by step solution

01

Determine the Ratio of Slit Separation to the Wavelength

The formula for the position of dark fringes in Young's double slit experiment is given by \(\sin(\theta) = m \lambda / d\), where m is the fringe order, \(\lambda\) is the wavelength, d is the slit separation, and \(\theta\) is the angle with respect to the original direction. Since the first completely dark fringes \(m = 1\) occur at \(\pm 19.0^{\circ}\), we can solve for the ratio of slit separation to the wavelength:\n \[d / \lambda = \sin (19^{\circ}) / 1\]
02

Determine the Intensity Ratio

The intensity of light at a given point in the interference pattern can be given by \(I/I_{max} = \frac{4\cos^2(\beta/2)}{1}\), where \(\beta = (2\pi / \lambda) d \sin (\theta)\). The minimum angle at which the intensity of the light is \(\frac{1}{10}\) the maximum intensity can be found by setting the intensity ratio to \(\frac{1}{10}\) and solving for \(\theta\). We do this as follows: \[\frac{1}{10} = \frac{4\cos^2(\beta/2)}{1}\] This can be solved to find \(\beta\), and consequently \(\theta\).
03

Calculate and Verify

Now we have two equations with known variables that we can solve to find the ratio of slit separation to the wavelength and the smallest angle with respect to the original direction.

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Key Concepts

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

Interference
The Young's Double Slit Experiment is the classic demonstration of light's wave nature. This phenomenon, known as "interference," happens when waves overlap and either strengthen each other (constructive interference) or weaken each other (destructive interference). When light waves from two slits meet, depending on the path difference, they create bright or dark bands on a screen called "fringes."

Here's how interference works in this context:
  • Constructive interference leads to bright fringes and occurs when the path difference between light waves is an integer multiple of the wavelength ewpage"), such as 0, 1, 2, etc.
  • Destructive interference results in dark fringes and occurs when the path difference is a half-integer multiple of the wavelength, such as 0.5位, 1.5位, etc.
In this exercise, we're concerned with the dark fringes (destructive interference). At the angle 卤19.0掳, the beam originally directed should cancel out, resulting in no visible light on the screen at this point. This happens because the waves from each slit destructively interfere.
Fringe Pattern
A "fringe pattern" is the sequence of dark and bright bands observed on the screen due to light interference. For Young's Double Slit Experiment, it appears as a series of alternating dark and bright lines. These patterns happen because light waves overlap differently at various points, resulting in constructive or destructive interference.

Here's what determines the fringe pattern:
  • The distance between the slits influences the spacing of the fringes.
  • The wavelength of light affects the fringe width. Shorter wavelengths mean fringes appear closer together.
  • The angle of observation also changes how the fringes spread on the screen.
In this scenario:
  • The first dark fringes at 卤19.0掳 are due to the path difference equaling one wavelength, causing complete destructive interference.
  • The smallest angle where the intensity is a tenth of the maximum shows how the intensity changes gradually across the fringes, not just at the peaks and troughs but smoothly varying based on the angle due to wave interference.
Intensity Ratio
The "intensity ratio" in an interference pattern describes how bright a fringe is compared to the maximum brightness observed. It's essential for understanding how the light's brightness varies across the interference pattern.

In Young's experiment:
  • The intensity is highest where all light waves from both slits constructively interfere completely, creating a bright fringe.
  • The intensity reduces where there's partial interference, leading to less bright or even dark fringes.
Given in the problem is how to find the angle where the intensity is only 10% (or \(\frac{1}{10}\)) of the maximum brightness. To find this:
  • Use the intensity formula \(I/I_{max} = \frac{4\cos^2(\beta/2)}{1}\), with \(\beta = (2\pi / \lambda) d \sin (\theta)\).
  • By setting \(I/I_{max} = \frac{1}{10}\), you can determine the corresponding angle for this specific intensity.
This approach lets you predict how much dimmer certain areas of the fringe pattern will be, providing insight into the intricate dance of light through slits.

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Most popular questions from this chapter

Red light with wavelength \(700 \mathrm{nm}\) is passed through a two-slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

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 \(\mathrm{TiO}_{2}\), which has an index of refraction of 2.62 , as the coating, what is the minimum film thickness that will cancel light of wavelength \(505 \mathrm{nm} ?\) (b) If this coating is too thin to stand up to wear, what other thickness would also work? Find only the three thinnest ones.

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

A compact disc (CD) is read from the bottom by a semiconductor laser with wavelength \(790 \mathrm{nm}\) passing through a plastic substrate of refractive index \(1.8 .\) When the beam encounters a pit, part of the beam is reflected from the pit and part from the flat region between the pits, so the two beams interfere with each other (Fig. E35.29). What must the minimum pit depth be so that the part of the beam reflected from a pit cancels the part of the beam reflected from the flat region? (It is this cancellation that allows the player to recognize the beginning and end of a pit.)

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 \(550 \mathrm{nm}\) 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?
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