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Chapter 28: Problem 73

A laser emits two wavelengths \(\left(\lambda_{1}=420 \mathrm{nm} ; \lambda_{2}=630 \mathrm{nm}\right)\). When these two wavelengths strike a grating with 450 lines/mm, they produce maxima (in different orders) that coincide. (a) What is the order \((m)\) of each of the two overlapping lines? (b) At what angle does this overlap occur?

Short Answer

Expert verified
Orders: 3 for 420 nm, 2 for 630 nm; Angle: 34.6掳.

Step by step solution

01

Understanding the Grating Equation

The grating equation is given by \( m \lambda = d \sin \theta \), where \( m \) is the order of the maximum, \( \lambda \) is the wavelength of light, \( d \) is the distance between adjacent grating lines, and \( \theta \) is the angle of the diffraction maximum. Given \( \lambda_1 = 420 \text{ nm} \) and \( \lambda_2 = 630 \text{ nm} \), we know \( d = \frac{1}{450 \text{ lines/mm}} = \frac{1}{450,000 \text{ lines/m}} \).
02

Set Up the Condition for Coinciding Maxima

For the maxima to coincide, they both must occur at the same angle \( \theta \). Thus, the same value of \( \sin \theta \) must satisfy the grating equation for both wavelengths. We equate:\[ m_1 \lambda_1 = m_2 \lambda_2 \].
03

Solve for the Order Ratio

Rearranging the equation from Step 2:\[ \frac{m_1}{m_2} = \frac{\lambda_2}{\lambda_1} = \frac{630 \text{ nm}}{420 \text{ nm}} = \frac{3}{2} \].This tells us that \( m_1 = 3 \) and \( m_2 = 2 \).
04

Calculate the Grating Line Spacing

Compute the grating line spacing \( d \):\[ d = \frac{1}{450,000} \approx 2.22 \times 10^{-6} \text{ m} \].
05

Substitute Back to Find the Angle

Using the first wavelength (\( \lambda_1 = 420 \text{ nm} \)) and its order (\( m_1 = 3 \)), substitute the values into the grating equation:\[ 3 \times 420 \times 10^{-9} = 2.22 \times 10^{-6} \sin \theta \].Solve for \( \sin \theta \): \[ \sin \theta = \frac{3 \times 420 \times 10^{-9}}{2.22 \times 10^{-6}} \approx 0.56757 \].
06

Calculate the Angle \( \theta \)

Now calculate \( \theta \) using \( \sin^{-1} \):\[ \theta = \sin^{-1}(0.56757) \approx 34.63^\circ \].

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Key Concepts

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

Wavelengths
Wavelengths are crucial in understanding the behavior of light, especially when dealing with phenomena like diffraction. Wavelength refers to the distance between consecutive peaks of a wave, such as light waves. In this particular exercise, two wavelengths are given: 420 nm and 630 nm. One nanometer (nm) is a billionth of a meter, making these wavelengths very small.
  • The 420 nm wavelength belongs to the violet spectrum of visible light.
  • The 630 nm wavelength belongs to the red spectrum.
These differences in wavelengths allow us to explore how light interacts with objects at the microscopic level, such as the grating in this problem. Different wavelengths diffract, or bend, differently due to their varying lengths. This variance is key in experiments involving diffraction gratings.
Order of Maximum
The order of maximum, represented by the symbol \( m \), is a major factor when working with diffraction patterns. The order of maximum indicates how many wavelengths fit into the path difference created by the grating, leading to a constructive interference where bright spots, or maxima, appear.In our original exercise, the order of maximums were \( m_1 = 3 \) for the 420 nm wavelength and \( m_2 = 2 \) for the 630 nm wavelength. These numbers tell us that:
  • Three full wavelengths of 420 nm occupy the same space as two of 630 nm.
  • This alignment results in the overlapping or coinciding of maxima for these orders, creating a bright spot noticed in diffraction patterns.
The integer number \( m \) highlights how many cycles of the wave fit in the setup, which is pivotal when aiming to achieve interference patterns like these.
Grating Equation
The grating equation forms the backbone of any problem involving diffraction gratings. It is represented as \( m \lambda = d \sin \theta \), where:
  • \( m \) is the order of maximum,
  • \( \lambda \) is the wavelength of light used,
  • \( d \) is the distance between adjacent grating lines,
  • \( \theta \) is the angle at which the maximum occurs.
For our problem, with 450 lines per mm, we have \( d = \frac{1}{450,000} \) meters. This formula relates the physical setup of the grating to the observable diffraction pattern. By substituting in the known quantities, we can solve for unknowns, like the angle \( \theta \), to understand how light behaves in the presence of the grating. Understanding this equation allows us to predict precisely where bright spots will appear on a screen observing the diffraction.
Laser Emission
Laser emission harnesses highly specific wavelengths of light to achieve highly focused beams. Lasers are remarkable for their coherent light 鈥� light with the same frequency and phase 鈥� which makes them quite useful in diffraction experiments. In the context of this exercise, a laser emitting two specific wavelengths, 420 nm and 630 nm, interacts with a diffraction grating. Here's why lasers are important in such setups:
  • They provide precise, monochromatic light needed to produce clear diffraction patterns.
  • The coherent nature ensures that interference patterns, like the ones whose maxima coincide, are easily identifiable.
Understanding how different wavelengths produced by a laser interact with objects like diffraction gratings helps in a myriad of applications, such as spectroscopy and even the creation of optical instruments.

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Most popular questions from this chapter

A two-slit experiment with red light produces a set of bright fringes. (a) Will the spacing between the fringes increase, decrease, or stay the same if the color of the light is changed to blue? (b) Choose the best explanation from among the following: I. The spacing between the fringes will increase because blue light has a greater frequency than red light. II. The fringe spacing decreases because blue light has a shorter wavelength than red light. III. Only the wave property of light is important in producing the fringes, not the color of the light. Therefore the spacing stays the same.

Images produced by structures within the eye (like lens fibers or cell fragments) are referred to as entoptic images. These images can sometimes take the form of "halos" around a bright light seen against a dark background. The halo in such a case is actually the bright outer rings of a circular diffraction pattern, like Figure \(28-21\), with the central bright spot not visible because it overlaps the direct image of the light. Find the diameter of the eye structure that causes a circular diffraction pattern with the first dark ring at an angle of \(3.7^{\circ}\) when viewed with monochromatic light of wavelength \(630 \mathrm{nm} .\) (Typical eye structures of this type have diameters on the order of \(10 \mu \mathrm{m}\). Also, the index of refraction of the vitreous humor is \(1.336 .\) )

(a) If a thin liquid film floating on water has an index of refraction less than that of water, will the film appear bright or dark in reflected light as its thickness goes to zero? (b) Choose the best explanation from among the following: I. The film will appear bright because as the thickness of the film goes to zero the phase difference for reflected rays goes to zero. II. The film will appear dark because there is a phase change at both interfaces, and this will cause destructive interference of the reflected rays.

Predict/Explain (a) In principle, do your eyes have greater resolution on a dark cloudy day or on a bright sunny day? (b) Choose the best explanation from among the following: I. Your eyes have greater resolution on a cloudy day because your pupils are open wider to allow more light to enter the eye. II. Your eyes have greater resolution on a sunny day because the bright light causes your pupil to narrow down to a a smaller opening.

Green light \((\lambda=546 \mathrm{nm})\) strikes a single slit at normal incidence. What width slit will produce a central maximum that is \(2.50 \mathrm{cm}\) wide on a screen \(1.60 \mathrm{m}\) from the slit?
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