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Chapter 26: Problem 34

\(\bullet\) A laser beam of unknown wavelength passes through a dif- fraction grating having 5510 lines/cm after striking it perpen- dicularly. Taking measurements, you find that the first pair of bright spots away from the central maximum occurs at \(\pm 15.4^{\circ}\) with respect to the original direction of the beam. (a) What is the wavelength of the light? (b) At what angle will the next pair of bright spots occur?

Short Answer

Expert verified
(a) Wavelength is 483 nm. (b) Next angle is 31.2掳.

Step by step solution

01

Identify the Formula

The formula for calculating the wavelength based on diffraction grating is given by the grating equation: \( d \sin\theta = m\lambda \), where \(d\) is the spacing between the grating lines, \(\theta\) is the angle of diffraction, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength of the light.
02

Calculate Grating Line Spacing

First, determine the spacing \(d\) between the lines in the diffraction grating. Given 5510 lines per cm, spacing \(d\) is calculated as: \( d = \frac{1}{5510}\) cm or \( d = \frac{1}{5510 \times 10^2}\) m.
03

Calculate Wavelength of Light (\(m=1\))

Using the grating equation for the first bright spot (\(m=1\)), we substitute \(d\), \(\theta=15.4^\circ\), and \(m=1\): \( \lambda = d \sin(15.4^\circ) \). Calculate \(\sin(15.4^\circ)\) and solve for \(\lambda\).
04

Calculate Sin of Angle \(15.4^\circ\)

Using a calculator, \(\sin(15.4^\circ) \approx 0.266\). Substitute this value into the equation obtained in the previous step.
05

Solve for Wavelength \(\lambda\)

Now calculate \(\lambda\) using \(\lambda = d \sin(15.4^\circ)\). Substituting values: \(\lambda = \left(\frac{1}{5510 \times 10^2}\right) \times 0.266\).
06

Wavelength Calculation

Complete the calculation to find \(\lambda\). \(\lambda = \left(\frac{1}{5510 \times 10^2}\right) \times 0.266\) results in \(\lambda \approx 4.83 \times 10^{-7}\) m or 483 nm.
07

Determine Angle for Next Pair of Bright Spots (\(m=2\))

For the next bright spot (\(m=2\)), use the equation \(d \sin\theta = 2\lambda\). Rearrange to find \(\theta\): \(\theta = \arcsin\left(\frac{2\lambda}{d}\right)\).
08

Calculate Sin of New Angle

Substitute \(\lambda = 483 \times 10^{-9}\) m into \(\theta = \arcsin\left(\frac{2 \times \lambda}{d}\right)\) and solve for \(\theta\).
09

Solve for Angle \(\theta\)

Calculate \(\theta = \arcsin\left( \frac{2 \times 483 \times 10^{-9}}{\frac{1}{5510 \times 10^2}} \right)\). After conducting this calculation, \(\theta \approx 31.2^\circ\).

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Key Concepts

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

Wavelength Calculation
In a diffraction grating setup, the wavelength of light can be calculated using a fundamental equation referred to as the grating equation. This equation is expressed as \(d \sin\theta = m\lambda\), where:
  • \(d\) represents the distance between adjacent lines on the grating, known as grating spacing.
  • \(\theta\) is the observed angle of diffraction for a bright spot.
  • \(m\) stands for the order of the diffraction maximum, typically a positive integer beginning with 1 for the first maximum.
  • \(\lambda\) is the wavelength of the light, which is what we're aiming to find.
To find the wavelength, follow these steps:
First, determine \(d\) by taking the reciprocal of the number of lines per cm, converting to meters. This gives us the spacing per line.
Next, for the first order diffraction maximum \((m=1)\), the angle \(\theta\) is measured as given (15.4掳 in this problem). Calculating \(\sin(\theta)\) and substituting all known values into the formula solves for the unknown \(\lambda\). In our example, performing the calculation results in a wavelength of approximately 483 nm.
Diffraction Angle
Diffraction grating experiments involve measuring the diffraction angle \(\theta\), which is crucial for calculating the properties of incoming light. This angle is the deviation from the original direction where bright spots, or maxima, appear.
To calculate \(\theta\), one measures the angle formed between the diffracted light and the original direction of the beam.
For higher orders of maxima, the diffraction angle changes due to the equation \(d \sin\theta = m\lambda\), where \(m\) denotes different orders.
  • For the first order \((m=1)\), the angle is typically smaller, representing the first set of bright spots observed.
  • Higher orders \((m=2, 3, ... )\) produce larger angles, representing subsequent bright spots further away from the central maximum.
Accurately measuring \(\theta\) is essential to determine wavelengths of light, making the understanding of its calculation a key component in a diffraction grating analysis.
Order of Maximum
The term "order of maximum," denoted by \(m\) in diffraction problems, refers to the series of bright spots observed when light passes through a diffraction grating.
These spots, also known as maxima, are the result of constructive interference where the path difference between successive rays equates to a multiple of the wavelength.
  • \(m=0\) refers to the central maximum 鈥� the brightest direct path.
  • \(m=1\) indicates the first order of maximum, the first pair of bright spots symmetrically located on either side of the central maximum.
  • \(m=2\) is the second order, farther away from the center, and so on.
The different orders are crucial for calculating where an angle will cause bright spots to appear, with each maximum鈥檚 position depending on its order in the formula \(d \sin\theta = m\lambda\). Solving for different orders helps predict the angles for these spots, such as the given 31.2掳 for the second order maximum in the exercise.

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Most popular questions from this chapter

\(\bullet\) A thin film of polystyrene of refractive index 1.49 is used as a nonreflecting coating for Fabulite (strontium titanate) of refractive index \(2.409 .\) What is the minimum thickness of the film required? Assume that the wavelength of the light in air is 480 \(\mathrm{nm}\) .

\(\bullet$$\bullet\) A uniform thin film of material of refractive index 1.40 coats a glass plate of refractive index \(1.55 .\) This film has the proper thickness to cancel normally incident light of wave- length 525 \(\mathrm{nm}\) that strikes the film surface from air, but it is somewhat greater than the minimum thickness to achieve this cancellation. As time goes by, the film wears away at a steady rate of 4.20 nm per year. What is the minimum number of years before the reflected light of this wavelength is now enhanced instead of cancelled?

\(\bullet\) A person with a radio-wave receiver starts out equidistant from two FM radio transmitters \(A\) and \(B\) that are 11.0 \(\mathrm{m}\) apart, each one emitting in-phase radio waves at 92.0 \(\mathrm{MH} z\) . She then walks so that she always remains 50.0 \(\mathrm{m}\) from transmitter \(B\) . See Fig- ure \(26.48 .\) ) For what values of \(x\) will she find the radio signal to be (a) maximally enhanced, (b) can- celled? Limit your solution to the cases where \(x \geq 50.0 \mathrm{m.}\)

\(\bullet$$\bullet\) Sensitive eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 \(\mathrm{nm}\) has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?

\(\bullet\) Nonglare glass. 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 thin 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 thicknesses would also work? Find only the three thinnest ones.
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