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The hydrogen spectrum has a red line at \(656 \mathrm{~nm}\) and a violet line at \(434 \mathrm{~nm}\). What angular separation between these two spectral lines is obtained with a diffraction grating that has 4500 lines/cm?

Short Answer

Expert verified
The answers for \(θ_{red}\), \(θ_{violet}\), and the angular separation will depend on the actual calculations completed in step 3.

Step by step solution

01

Interpret the Data

Identify the given data. From the problem, we have the following information: The wavelength of the red line (\(λ_{red}\)) is 656 nm or \(656 \times 10^{-9} \) meters. The wavelength of the violet line (\(λ_{violet}\)) is 434 nm or \(434 \times 10^{-9} \) meters. The grating spacing (d) can be determined by taking the reciprocal of the grating's given lines per cm, converting it to meters: \(d = 1/(4500 \times 10) = 2.22 \times 10^{-5}\) meters.
02

Apply Diffraction Grating Formula

The condition for constructive interference, according to the diffraction grating formula, is \(d \cdot sin(θ) = m \cdot λ\), where θ is the angle of diffracted light, m is the order of diffraction (assuming to be 1 in our case since it's not given), d is the distance between the slits, and λ is the wavelength of light. Solve the formula for θ, to find the angles for both red and violet spectral lines: For red line:\(sin(θ_{red}) = λ_{red} / d\), hence\(θ_{red} = arcsin( λ_{red} / d ) \).For violet line: \(sin(θ_{violet}) = λ_{violet} / d\), hence\(θ_{violet} = arcsin( λ_{violet} / d ) \).
03

Find Angular Separation

The angular separation can be obtained by subtracting the two angles. Therefore, Angular separation = \(θ_{red} - θ_{violet}\). Plug the respective values into the formula to compute the answer.

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Key Concepts

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

Spectral Lines Angular Separation
The concept of spectral lines angular separation is central to understanding how a diffraction grating can be used to differentiate between different wavelengths of light in a spectrum. Consider the hydrogen spectrum, which consists of distinct colored lines, including a red line at 656 nm and a violet line at 434 nm.
With this concept, it becomes critical for students and scientists alike to measure how far apart these lines appear when passed through a diffraction grating. In a practical scenario, the angular separation is the physical angle between the diffracted light of different wavelengths.
This angle becomes increasingly important in applications such as spectroscopy, where determining the composition of a light source is essential. To accurately find the angular separation, one must understand how to apply the diffraction grating formula and infer the angles at which different wavelengths of light are diffracted.
Diffraction Grating Formula
The diffraction grating formula is a fundamental aspect of wave optics and is expressed as:\[d \cdot \sin(\theta) = m \cdot \lambda\]. Here, \(d\) represents the distance between adjacent slits (grating spacing), \(\theta\) is the angle of diffracted light, \(m\) is the order of diffraction, usually an integer that represents the series of maxima produced by the grating, and \(\lambda\) is the wavelength of the light being examined.

How the Formula Is Applied

In the case of the hydrogen spectrum problem, the grating spacing (
\(d\)) is calculated based on the number of lines per unit length, which in this case is 4500 lines per centimeter. Converting this value to meters gives us the precise spacing needed to utilize the diffraction grating formula to calculate the angle for each wavelength.
  • The angle for red light is calculated using the wavelength of 656 nm.
  • The angle for violet light uses the wavelength of 434 nm.
By solving the equation for each wavelength, we obtain the individual angles, which can then be used to find the angular separation. The formula's simplicity makes it a powerful tool for analyzing the diffraction patterns to determine various light properties.
Constructive Interference
Constructive interference is a wave phenomenon that occurs when two or more waves combine to produce a wave with a larger amplitude. In the context of a diffraction grating, constructive interference is required to produce the bright spectral lines that are characteristic of specific elements, like hydrogen in our example.

Creating Bright Lines

This constructive interference happens at specific angles where the path difference between light from adjacent slits is an integer multiple of the wavelength, satisfying the grating formula:\[d \cdot \sin(\theta) = m \cdot \lambda\]. For a given order of interference (
\(m\)), maximum brightness occurs at certain angles where this constructive interference is maximized.
Students must grasp that it's the precise alignment of the wave crests from different slits that leads to the intensified light—a key aspect of understanding how diffraction gratings dissect light into its composite wavelengths. The enhanced understanding of constructive interference not only explains the patterns seen in diffraction but also contributes to our understanding of the wave nature of light.

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Most popular questions from this chapter

Nonreflective coatings on camera lenses reduce the loss of light at the surfaces of multilens systems and prevent internal reflections that might mar the image. Find the minimum thickness of a layer of magnesium fluoride \((n=1.38)\) on flint glass \((n=1.66)\) that will cause destructive interference of reflected light of wavelength \(550 \mathrm{~nm}\) near the middle of the visible spectrum.

A soap bubble \((n=1.93)\) having a wall thickness of \(120 \mathrm{~nm}\) is floating in air. (a) What is the wavelength of the visible light that is most strongly reflected? (b) Explain how a bubble of different thickness could also strongly reflect light of this same wavelength. (c) Find the two smallest film thicknesses larger than the one given that can produce strongly reflected light of this same wavelength.

Light with a wavelength in vacuum of \(546.1 \mathrm{~nm}\) falls perpendicularly on a biological specimen that is \(1.000 \mu \mathrm{m}\) thick. The light splits into two beams polarized at right angles, for which the indices of refraction are \(1.320\) and \(1.333\), respectively. (a) Calculate the wavelength of each component of the light while it is traversing the specimen. (b) Calculate the phase difference between the two beams when they emerge from the specimen.

In a Young's interference experiment, the two slits are separated by \(0.150 \mathrm{~mm}\) and the incident light includes two wavelengths: \(\lambda_{1}=540 \mathrm{~nm}\) (green) and \(\lambda_{2}=\) \(450 \mathrm{~nm}\) (blue). The overlapping interference patterns are observed on a screen \(1.40 \mathrm{~m}\) from the slits. (a) Find a relationship between the orders \(m_{1}\) and \(m_{2}\) that determines where a bright fringe of the green light coincides with a bright fringe of the blue light. (The order \(m_{1}\) is associated with \(\lambda_{1}\), and \(m_{2}\) is associated with \(\lambda_{2} .\) (b) Find the minimum values of \(m_{1}\) and \(m_{2}\) such that the overlapping of the bright fringes will occur and find the position of the overlap on the screen.

A diffraction grating has \(4.200 \times 10^{3}\) rulings per centimeter. The screen is \(2.000 \mathrm{~m}\) from the grating. In parts (a) through (e), round each result to four digits, using the rounded values for subsequent calculations. (a) Compute the value of \(d\), the distance between adjacent rulings. Express the answer in meters. (b) Calculate the angle of the second-order maximum made by the \(589.0-\mathrm{nm}\) wavelength of sodium. (c) Find the position of the secondary maximum on the screen. (d) Repeat parts (b) and (c) for the second-order maximum of the \(589.6-\mathrm{nm}\) wavelength of sodium, finding the angle and position on the screen. (e) What is the distance between the two secondary maxima on the screen? (f) Now find the distance between the two secondary maxima without rounding, carrying all digits in calculator memory. How many significant digits are in agreement? What can you conclude about rounding and the use of intermediate answers in this case?

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