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Chapter 27: Problem 43

A diffraction grating is \(1.50 \mathrm{cm}\) wide and contains 2400 lines. When used with light of a certain wavelength, a third-order maximum is formed at an angle of \(18.0^{\circ} .\) What is the wavelength (in \(\mathrm{nm}\) )?

Short Answer

Expert verified
The wavelength of the light is approximately 644 nm.

Step by step solution

01

Understand the Problem

We need to find the wavelength of light that causes a third-order diffraction maximum at an angle of \(18.0^{\circ}\). We have a diffraction grating that is \(1.50\,\text{cm}\) wide and contains \(2400\) lines.
02

Calculate Grating Spacing

First, calculate the distance between adjacent lines (grating spacing), denoted as \(d\). Since there are \(2400\) lines in \(1.50\,\text{cm}\), the grating spacing \(d\) is:\[d = \frac{1.50\,\text{cm}}{2400}\m = 0.0625\,\mathrm{mm} = 6250\,\mathrm{nm}\]
03

Apply Diffraction Grating Equation

Use the diffraction grating formula for maxima:\[m \cdot \lambda = d \cdot \sin(\theta)\]where \(m\) is the order of the maximum (\(m=3\)), \(\lambda\) is the wavelength, \(d\) is the grating spacing, and \(\theta\) is the angle (\(18.0^{\circ}\)).
04

Solve for Wavelength

Substitute the known values into the formula:\[3 \cdot \lambda = 6250\,\mathrm{nm} \cdot \sin(18.0^{\circ})\]Calculate \(\sin(18.0^{\circ}) \approx 0.309\):\[3 \cdot \lambda = 6250 \cdot 0.309 = 1931.25\,\mathrm{nm}\]Solve for \(\lambda\):\[\lambda = \frac{1931.25\,\mathrm{nm}}{3} \approx 643.75\,\mathrm{nm}\]
05

Final Step: Round the Result

Round the wavelength to a sensible number of decimal places. The wavelength is approximately \(644\,\mathrm{nm}\).

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Key Concepts

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

Wavelength Calculation
To find the wavelength of light that causes a diffraction pattern, we use the diffraction grating formula:
  • \( m \cdot \lambda = d \cdot \sin(\theta) \)
  • Where \( m \) is the order of the maximum.
  • \( \lambda \) is the wavelength (what we're solving for).
  • \( d \) is the grating spacing (distance between lines).
  • \( \theta \) is the angle of the maximum.
In the given exercise, we're focusing on a third-order maximum, where \( m = 3 \), and \( \theta = 18.0^{\circ} \). Plug these values into the formula to find \( \lambda \).
Once the known values are substituted, solving the equation gives the wavelength of approximately \( 644 \ \text{nm} \). This result tells us the specific color characteristic of light, enabling various scientific and technological applications.
Third-Order Maximum
The concept of a "third-order maximum" refers to the third bright spot in the diffraction pattern. These maxima occur at specific angles determined by the relationship between the light's wavelength and the grating spacing.
In this context, the equation \( m \cdot \lambda = d \cdot \sin(\theta) \) is used, with \( m = 3 \), to determine the angle at which this third bright fringe appears.
  • Each order corresponds to an integer multiple of wavelengths fitting the spacing.
  • Higher orders (like the third order) typically appear at larger angles.
This third-order maximum signifies not just a mathematical solution but a real-world observation in experiments involving diffraction gratings.
Grating Spacing
Grating spacing, or the distance between adjacent lines on a grating, is a critical factor in diffraction experiments. It essentially determines where different maxima will appear.
  • Calculated as: \( d = \frac{L}{N} \), where \( L \) is the total width of the grating, and \( N \) is the number of lines.
  • In the provided problem, \( d = \frac{1.50 \ \text{cm}}{2400} = 6250 \ \text{nm} \).
The grating spacing \( d \) influences how wavelengths interact to produce maxima.
Smaller spacings lead to greater angular spread between maxima, helping determine the resolution of the diffraction pattern. Understanding this concept is vital for interpreting the structures seen in spectra and other optical analyses.

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Most popular questions from this chapter

Light that has a wavelength of 668 nm passes through a slit \(6.73 \times 10^{-6} \mathrm{m}\) wide and falls on a screen that is \(1.85 \mathrm{m}\) away. What is the distance on the screen from the center of the central bright fringe to the third dark fringe on either side?

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is \(296 \mathrm{nm}\). What is the vacuum wavelength of the light in \(\mathrm{nm} ?\)

There are 5620 lines per centimeter in a grating that is used with light whose wavelength is \(471 \mathrm{nm} .\) A flat observation screen is located at a distance of \(0.750 \mathrm{m}\) from the grating. What is the minimum width that the screen must have so the centers of all the principal maxima formed on either side of the central maximum fall on the screen?

Late one night on a highway, a car speeds by you and fades into the distance. Under these conditions the pupils of your eyes have diameters of about \(7.0 \mathrm{mm} .\) The taillights of this car are separated by a distance of \(1.2 \mathrm{m}\) and emit red light (wavelength \(=660 \mathrm{nm}\) in vacuum). How far away from you is this car when its taillights appear to merge into a single spot of light because of the effects of diffraction?

It is claimed that some professional baseball players can see which way the ball is spinning as it travels toward home plate. One way to judge this claim is to estimate the distance at which a batter can first hope to resolve two points on opposite sides of a baseball, which has a diameter of \(0.0738 \mathrm{m} .\) (a) Estimate this distance, assuming that the pupil of the eye has a diameter of \(2.0 \mathrm{mm}\) and the wavelength of the light is \(550 \mathrm{nm}\) in vacuum. (b) Considering that the distance between the pitcher's mound and home plate is \(18.4 \mathrm{m},\) can you rule out the claim based on your answer to part (a)?
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