/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 A diffraction grating with slits... [FREE SOLUTION] | 91Ó°ÊÓ

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A diffraction grating with slits \(0.60 \times 10^{-3} \mathrm{cm}\) apart is illuminated by light with a wavelength of \(500 \mathrm{nm}\). At what angle will the third-order maximum appear?

Short Answer

Expert verified
The third-order maximum appears at approximately 1.43°.

Step by step solution

01

Convert Units

First, convert all units to meters. The slit separation, provided as \(0.60 \times 10^{-3} \mathrm{cm}\), should be converted to meters. Since \(1 \, cm = 10^{-2} \, m\), we have: \(d = 0.60 \times 10^{-3} \times 10^{-2} = 0.60 \times 10^{-5} \, m\). The wavelength is given as \(500 \, nm\), which is \(500 \times 10^{-9} \, m\).
02

Use the Diffraction Grating Equation

The diffraction grating equation is \(d \sin(\theta) = m \lambda\), where \(d\) is the distance between slits, \(\theta\) is the diffraction angle, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength. Substitute the known values: \(d = 0.60 \times 10^{-5} \, m\), \(\lambda = 500 \times 10^{-9} \, m\), and \(m = 3\) (since we are finding the third-order maximum).
03

Rearrange the Equation

Rearrange the diffraction grating equation to solve for \(\theta\). This gives \(\sin(\theta) = \frac{m \lambda}{d}\). Substituting the values: \(\sin(\theta) = \frac{3 \times 500 \times 10^{-9}}{0.60 \times 10^{-5}}\).
04

Calculate \(\sin(\theta)\)

Perform the calculation for \(\sin(\theta)\): \(\sin(\theta) = \frac{1500 \times 10^{-9}}{0.60 \times 10^{-5}} = \frac{1500}{60000}\). Simplifying gives \(\sin(\theta) = 0.025\).
05

Solve for \(\theta\)

Find \(\theta\) by taking the inverse sine: \(\theta = \sin^{-1}(0.025)\). Use a calculator to compute this value, giving \(\theta \approx 1.43^\circ\).

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Key Concepts

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

Wavelength
Wavelength is a key concept in understanding light behavior, particularly in phenomena like diffraction. Imagine it as the distance between consecutive peaks in a wave, much like waves on the sea. It is often denoted by the symbol \( \lambda \). In the context of light, wavelengths are usually measured in nanometers (nm), where 1 nm = \(10^{-9}\) meters.
Understanding wavelength is crucial because it determines the color of light we see. For example:
  • Wavelengths around 400 nm appear violet.
  • Wavelengths around 700 nm appear red.
In our exercise, the light used has a wavelength of 500 nm, which places it in the middle of the visible spectrum, giving it a greenish hue. Wavelength not only influences color but is also critical in calculations involving diffraction, as it interacts with the slit spacing in a diffraction grating to create interference patterns.
Diffraction Angle
The diffraction angle plays a crucial role in diffraction grating experiments. It is the angle at which light is deviated due to passing through the grating. Think of it like aiming a flashlight through a picket fence and noticing the light beams' spread.
Mathematically, the angle \( \theta \) for a maximum is calculated using the equation:\[ d \sin(\theta) = m \lambda \]where \( d \) is the slit spacing, \( m \) is the order of maximum, and \( \lambda \) is the wavelength.
For practical purposes, solving for the angle involves rearranging the expression to find \( \sin(\theta) \). You can then find \( \theta \) using the inverse sine function. This gives us the exact angle where light will appear brightest for each order of maximum. In our example, the third order results in a diffraction angle of approximately 1.43 degrees, suggesting a slight deviation due to the slit width and light wavelength.
Order of Maximum
The order of maximum is a term used in diffraction grating calculations to indicate the sequence of bright spots, or maxima, that appear due to interference. It is represented by \( m \) in the diffraction equation.
The first-order maximum (\( m = 1 \)) is the initial main bright spot either side of the central one. The second order (\( m = 2 \)) comes next, and so on. Each maximum corresponds to light waves being perfectly in phase, adding up to a bright spot at these specific angles.
In our calculation, we determined the third-order maximum (\( m = 3 \)), which tells us this is the third bright spot on either side of the central maximum. As the order increases, it means the angle is further from the normal line (or center position), and the positions of the maxima are spread further apart. This concept helps scientists measure light properties very accurately.

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Most popular questions from this chapter

Consider the Fraunhofer diffraction pattern for eight very narrow parallel slits under monochromatic illumination. (a) Sketch the resulting irradiance distribution. (b) Explain why the first minimum occurs, from a phasor perspective. (c) Why is the electric field zero midway between principal peaks? (d) What does the phasor diagram of the field amplitude look like for the second minimum (measured from the zeroth principal maximum)? (e) What are the angles between successive phasors at each minimum considered above?

Using Rayleigh's criterion, determine the smallest angle subtended by two points of equal brightness that can just be resolved by the human eye. Assume a pupil diameter of \(2.0 \mathrm{mm}\) and a mean wavelength of \(550 \mathrm{nm}\). The index of refraction of the medium within the eye is 1.337

A grating has a total width of \(10.0 \mathrm{cm}\) and contains 600 lines/ \(\mathrm{mm} .\) What is its resolving power in the second-order spectrum? At a mean wavelength of \(540 \mathrm{nm}\), what wavelength difference can it resolve?

A transmission grating whose lines are separated by \(3.0 \times\) \(10^{-6} \mathrm{m}\) is illuminated by a narrow beam of red light \(\left(\lambda_{0}=694.3 \mathrm{nm}\right)\) from a ruby laser. Spots of diffracted light, on both sides of the undeflected beam, appear on a screen \(2.0 \mathrm{m}\) away. How far from the central axis is each of the two nearest spots?

Plane waves from a magnesium lamp \((\lambda=518.36 \mathrm{nm})\) arrive perpendicularly on an opaque screen containing a long 0.250 -mm-wide slit. A large nearby positive lens forms a sharp image of the Fraunhofer diffraction pattern on a screen. The center of the fourth dark fringe is found to be \(1.20 \mathrm{mm}\) from the central axis. Determine the focal length of the lens.

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