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With light from a gaseous discharge tube incident normally on a grating with slit separation \(1.73 \mu \mathrm{m}\), sharp maxima of green light are experimentally found at angles \(\theta=\pm 17.6^{\circ}, 37.3^{\circ},-37.1^{\circ}\), \(65.2^{\circ}\), and \(-65.0^{\circ} .\) Compute the wavelength of the green light that best fits these data.

Short Answer

Expert verified
The wavelength of the green light is approximately 523.3 nm.

Step by step solution

01

Understand the Grating Equation

The grating equation is given by \(d \sin \theta = m \lambda\), where \(d\) is the slit separation, \(\theta\) is the angle at which the maxima occur, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength of light. We are given \(d = 1.73\,\mu\mathrm{m}\), and the angles \(\theta = \pm 17.6^{\circ}, 37.3^{\circ}, -37.1^{\circ}, 65.2^{\circ}, -65.0^{\circ}\). Our goal is to find \(\lambda\), the wavelength of green light.
02

Identifying the Order of Peaks

The appearance of maxima at similar angles (\(\theta = 37.3^{\circ}\) and \(\theta = -37.1^{\circ}, 65.2^{\circ}\) and \(\theta = -65.0^{\circ}\)) suggests that they are higher order maxima. We need to decide which angles correspond to consecutive orders. Let \(\theta_1 = 17.6^{\circ}\) correspond to the first order \(m=1\), then \(\theta_2 = 37.3^{\circ}\) for \(m=2\), and \(\theta_3 = 65.2^{\circ}\) for \(m=3\). The negative angles give symmetrical results for corresponding maximas.
03

Calculate Wavelength for Each Order

Apply the grating equation to find \(\lambda\) for each order:- For \(m=1\) and \(\theta_1 = 17.6^{\circ}\): \[ \lambda_1 = \frac{d \sin \theta_1}{1} = 1.73 \times 10^{-6} \sin(17.6^{\circ}) = 1.73 \times 10^{-6} \times 0.302\,293 = 0.523\,\mu\mathrm{m} \]- For \(m=2\) and \(\theta_2 = 37.3^{\circ}\): \[ \lambda_2 = \frac{d \sin \theta_2}{2} = \frac{1.73 \times 10^{-6} \sin(37.3^{\circ})}{2} = 1.73 \times 10^{-6} \times 0.607\,930 / 2 = 0.525\,\mu\mathrm{m} \]- For \(m=3\) and \(\theta_3 = 65.2^{\circ}\): \[ \lambda_3 = \frac{d \sin \theta_3}{3} = \frac{1.73 \times 10^{-6} \sin(65.2^{\circ})}{3} = 1.73 \times 10^{-6} \times 0.906\,308 / 3 = 0.522\,\mu\mathrm{m} \]
04

Average Wavelength Calculation

The wavelength of green light is best represented by the average of these wavelengths. Calculate the average:\[ \lambda_{\text{avg}} = \frac{0.523\, + 0.525\, + 0.522\,}{3} \mu\mathrm{m} = 0.5233\,\mu\mathrm{m} \]
05

Convert Wavelength to Nanometers

Convert the average wavelength from micrometers to nanometers for a standard notation of light wavelength:\[ \lambda_{\text{avg}} = 0.5233 \times 1000 = 523.3\,\mathrm{nm} \]

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Key Concepts

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

Wavelength Calculation
When calculating the wavelength of light using a diffraction grating, we rely on the grating equation to guide us. The grating equation is expressed as \(d \sin \theta = m \lambda\). Here, \(m\) is the order of the diffraction maximum, \(d\) is the slit separation, \(\theta\) is the diffraction angle, and \(\lambda\) is the wavelength of the light. In this exercise, \(d\) is given as \(1.73 \mu \mathrm{m}\), and several angles \(\theta\) are provided for calculation.
To find the wavelength of green light from these angles, first identify the angles that correspond to the diffraction maxima. Then use the grating equation to solve for \(\lambda\) by substituting \(d\) and the appropriate values of \(\theta\) and \(m\). Each calculation yields a wavelength for a specific order of diffraction. Averaging these wavelengths gives the best estimation of the light's true wavelength. Finally, convert the average wavelength from micrometers to nanometers for a more common representation.
Order of Maxima
When light passes through a diffraction grating, it forms bright spots known as maxima at certain angles. These angles depend on the order of maxima, represented by \(m\) in the grating equation. The order is crucial as it determines the position of these bright spots and is an integer value starting from 1.
In this exercise, the angles \(\theta = \pm 17.6^{\circ}\) correspond to first order maxima \(m=1\). Similarly, \(37.3^{\circ}\) and \(-37.1^{\circ}\) are second order maxima with \(m=2\), while \(65.2^{\circ}\) and \(-65.0^{\circ}\) represent third order maxima \(m=3\). It's important to identify these correctly because the order affects the calculation of the wavelength of the light.
  • The first order maxima give us the initial reference to begin calculations.
  • Higher-order maxima allow verification of consistency across different values of \(m\).
Each order provides a piece of the puzzle to solve for the correct wavelength of light.
Grating Equation
The grating equation \(d \sin \theta = m \lambda\) is the cornerstone of solving diffraction problems. It links the physical separation of the grating's slits \(d\), the angle of diffraction \(\theta\), the order of maxima \(m\), and the wavelength \(\lambda\).
This equation allows us to harness the known values to determine the unknown variable, usually the wavelength of the light in question. By inputting the angle for each observed maxima and its corresponding order, we calculate distinct values of \(\lambda\) for each scenario.
  • Use given slit separation: \(d = 1.73 \mu \mathrm{m}\).
  • Apply correct \(\theta\) values for each maxima order.
  • Compute \(\lambda\) for multiple orders to ensure consistency.
Repeated calculations through the grating equation reaffirm the reliability of the results, ultimately averaging the values for a final output.
Light Diffraction
Diffraction occurs when waves, such as light, encounter an obstacle or opening that causes them to bend or spread. When light passes through a grating, a unique pattern of light and dark bands forms, showing areas of constructive and destructive interference known as diffraction maxima and minima.
The appearance of these maxima at specific angles provides crucial information about the light's wavelength. Green light, used in this exercise, diffracts and forms sharp maxima at particular angles when it interacts with the grating. This spread allows us to measure and calculate the unknown wavelength with accuracy.
  • Diffraction patterns are produced due to interference of light waves.
  • Measured angles of maxima help derive wavelengths using the grating equation.
  • Understanding diffraction expands both theoretical and practical comprehension of wave behavior.
Light diffraction is a gateway into deeper insights of wave interactions and is widely applicable in understanding optics and spectroscopy.

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Most popular questions from this chapter

A 0.10-mm-wide slit is illuminated by light of wavelength 589 \(\mathrm{nm}\). Consider a point \(P\) on a viewing screen on which the diffraction pattern of the slit is viewed; the point is at \(30^{\circ}\) from the central axis of the slit. What is the phase difference between the Huygens wavelets arriving at point \(P\) from the top and midpoint of the slit?

The two headlights of an approaching automobile are \(1.4 \mathrm{~m}\) apart. At what (a) angular separation and (b) maximum distance will the eye resolve them? Assume that the pupil diameter is \(5.0 \mathrm{~mm}\), and use a wavelength of \(550 \mathrm{~nm}\) for the light. Also assume that diffraction effects alone limit the resolution so that Rayleigh's criterion can be applied.

A source containing a mixture of hydrogen and deuterium atoms emits red light at two wavelengths whose mean is \(656.3 \mathrm{~nm}\) and whose separation is \(0.180 \mathrm{~nm} .\) Find the minimum number of lines needed in a diffraction grating that can resolve these lines in the first order.

Nuclear-pumped x-ray lasers are seen as a possible weapon to destroy ICBM booster rockets at ranges up to \(2000 \mathrm{~km}\). One limitation on such a device is the spreading of the beam due to diffraction, with resulting dilution of beam intensity. Consider such a laser operating at a wavelength of \(1.40 \mathrm{~nm} .\) The element that emits light is the end of a wire with diameter \(0.200 \mathrm{~mm}\). (a) Calculate the diameter of the central beam at a target \(2000 \mathrm{~km}\) away from the beam source. (b) By what factor is the beam intensity reduced in transit to the target? (The laser is fired from space, so that atmospheric absorption can be ignored.)

In a single-slit diffraction experiment, there is a minimum of intensity for orange light \((\lambda=600 \mathrm{~nm})\) and a minimum of intensity for blue- green light \((\lambda=500 \mathrm{~nm})\) at the same angle of \(1.00\) mrad. For what minimum slit width is this possible?

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