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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 is 643.75 nm.

Step by step solution

01

Determine the Grating Spacing

First, determine the grating spacing or separation, denoted as \(d\). Since the grating has 2400 lines over a width of \(1.50 \, \mathrm{cm}\), we can convert this width to meters: \(1.50 \, \mathrm{cm} = 0.015 \, \mathrm{m}\). Then, divide the width by the number of lines to find \(d\):\[d = \frac{0.015 \, \mathrm{m}}{2400} = 6.25 \times 10^{-6} \, \mathrm{m}\]
02

Understand the Diffraction Condition

The diffraction condition for the maxima is given by the formula:\[m \lambda = d \sin \theta\]Where \(m\) is the order of the maximum, \(\lambda\) is the wavelength, \(d\) is the grating spacing, and \(\theta\) is the diffraction angle. Here, \(m = 3\) for the third-order maximum, and \(\theta = 18.0^{\circ}\).
03

Calculate the Wavelength

Substitute the known values into the diffraction condition and solve for the wavelength \(\lambda\):\[\lambda = \frac{d \sin \theta}{m}\]Substituting the given values:\[\lambda = \frac{6.25 \times 10^{-6} \, \mathrm{m} \times \sin(18^{\circ})}{3}\]
04

Solve for the Wavelength

Calculate \(\sin(18^{\circ})\):\[\sin(18^{\circ}) \approx 0.309\]Now substitute this value back into the equation:\[\lambda = \frac{6.25 \times 10^{-6} \, \mathrm{m} \times 0.309}{3} = 6.4375 \times 10^{-7} \, \mathrm{m}\]
05

Convert Wavelength to Nanometers

Convert the wavelength from meters to nanometers by multiplying by \(10^9\) (since \(1 \, \mathrm{m} = 10^9 \, \mathrm{nm}\)):\[\lambda = 6.4375 \times 10^{-7} \, \mathrm{m} \times 10^9 \, \mathrm{nm/m} = 643.75 \, \mathrm{nm} \]

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Key Concepts

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

Grating Spacing
In a diffraction grating setup, the grating spacing, denoted as \(d\), is a key component in determining how light diffuses into different angles. Consider a grating that is 1.50 cm wide with 2400 parallel lines. Grating spacing indicates the distance between consecutive lines, and it is calculated by dividing the total width by the number of lines. Here, the width in meters is given by:
  • 1.50 cm converted to meters: 0.015 m
  • Number of lines: 2400
The formula to find the grating spacing is: \[d = \frac{0.015 \, \mathrm{m}}{2400} = 6.25 \times 10^{-6} \, \mathrm{m}\] This value of \(d\) determines how the incident light will be spread out. Smaller spacings result in larger diffraction angles, helping to identify different wavelengths precisely.
Diffraction Condition
The diffraction condition governs the formation of maxima (constructive interference) in a diffraction pattern and is given by the equation:\[m \lambda = d \sin \theta\]
  • \(m\) is the order of maximum, representing the number of wavelengths that fit into the path difference between adjacent slits.
  • \(\lambda\) is the wavelength of the light being used.
  • \(d\) is the grating spacing.
  • \(\theta\) is the angle of diffraction, the angle at which the maxima are observed.
In this problem, the diffraction condition allows us to solve for the wavelength \(\lambda\) when other variables are known. Here, we know it is a third-order maximum, meaning \(m = 3\), with a diffraction angle of \(18.0^\circ\) and a previously determined \(d\). Understanding this condition is essential for figuring out how different wavelengths will appear at different angles, leading to the observation of color separation.
Wavelength Calculation
Calculating the wavelength in a diffraction grating experiment involves the application of the diffraction condition formula. To find the wavelength \(\lambda\), we rearrange the formula:\[\lambda = \frac{d \sin \theta}{m}\]Plug in the given and calculated values:
  • \(d = 6.25 \times 10^{-6} \, \mathrm{m}\)
  • \(\theta = 18.0^\circ\)
  • \(m = 3\)
First, compute \(\sin(18^\circ)\) which is approximately 0.309:\[\lambda = \frac{6.25 \times 10^{-6} \, \mathrm{m} \times 0.309}{3}\]The calculated wavelength is:\[\lambda \approx 6.4375 \times 10^{-7} \, \mathrm{m}\]This is how light wavelengths are measured in such experiments, demonstrating the use of trigonometric functions to find unknowns in wave optics.
Third-order Maximum
When studying diffraction patterns, maxima are labeled based on the order of their occurrence—first, second, third, and so on. A third-order maximum refers to the third instance of constructive interference in the pattern. The order \(m = 3\) in this scenario indicates that three full wavelengths fit into the path difference created by the grating.
The diffraction condition \(m \lambda = d \sin \theta\) plays a crucial role in determining the wavelength associated with each order. Each higher order involves wider angles for their maxima. This relationship helps in precisely characterizing the spectral components of the light source. It's the same principle that allows complex diffractive analyses in scientific research and industry applications. The higher the order, the more separation there is between individual components in the resulting spectrum.

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Most popular questions from this chapter

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.

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?

The dark fringe for \(m=0\) in a Young's double-slit experiment is located at an angle of \(\theta=15^{\circ} .\) What is the angle that locates the dark fringe for \(m=1 ?\)

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 diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and \(660 \mathrm{~nm}\). What is (are) the wavelength(s) of the incident light that could have produced this maximum?

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