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Chapter 36: Problem 26

If a diffraction grating produces a third-order bright spot for red light (of wavelength 700 nm) at 65.0\(^\circ\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength 400 nm)?

Short Answer

Expert verified
The second-order bright spot for violet light is at 29.6°.

Step by step solution

01

Understand the Diffraction Grating Equation

The diffraction grating equation is given by \(d \sin \theta = m \lambda\), where \(d\) is the distance between grating lines, \(\theta\) is the angle of the bright spot, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength of the light.
02

Calculate Grating Line Density Using Red Light

For the third-order bright spot with red light (\(\lambda = 700\) nm) at \(\theta = 65.0^\circ\): Rearrange the equation to solve for \(d\): \[d = \frac{m \lambda}{\sin \theta} = \frac{3 \times 700 \text{ nm}}{\sin 65.0^\circ}\].Calculate \(d\).
03

Use Grating Line Density for Violet Light

Use the same \(d\) calculated for violet light (\(\lambda = 400\) nm) for the second-order maximum: \(m = 2\). Rearrange the equation to solve for \(\theta\) for violet light: \[\sin \theta = \frac{m \lambda}{d} = \frac{2 \times 400 \text{ nm}}{d}\]. Calculate \(\theta\) using the \(d\) from the previous step.
04

Calculate and Verify the Angle for Violet Light

Substitute the value of \(d\) obtained in Step 2 into the equation from Step 3. Solve for \(\theta\): \[\theta = \arcsin \left( \frac{2 \times 400}{d} \right)\]. Calculate \(\theta\) to find the angle for the second-order bright spot.

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Key Concepts

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

Wavelength
In the world of physics, **wavelength** is a fundamental concept that describes the distance between consecutive peaks of a wave, especially in a periodic wave. Waves can be found in various forms such as sound, water, and electromagnetic waves—like light. For light waves, wavelength determines color, with red having the longest wavelength visible to the human eye and violet having one of the shortest. Wavelength is typically measured in nanometers (nm), with 1 nm equaling one billionth of a meter. In the given problem involving light diffraction through a grating, the red light has a wavelength of 700 nm, while the violet light is 400 nm. Understanding wavelength helps in predicting how waves interact when encountering obstacles or openings, such as occurring in diffraction. Longer wavelengths, like red light, will spread out more than shorter wavelengths like violet when passing through a medium or around a barrier.
Third-Order Bright Spot
Bright spots, also known as maxima, occur due to the constructive interference of light waves. A **third-order bright spot** refers to the third position of maximum brightness on either side of the central maximum. This occurs when the path difference between waves equals three whole wavelengths, leading to constructive interference. In the context of using a diffraction grating, different wavelengths will result in bright spots at different angles. For red light in this problem, the third-order bright spot occurs at an angle of 65.0°. This means the pattern for red light has extended three full wavelengths from the central line to reach this point of intensity. This concept allows scientists and engineers to measure and analyze the properties of light by observing these diffracted patterns. Each order of maxima can provide insights into the wavelength of the light used.
Diffraction Grating Equation
The **diffraction grating equation** is essential in understanding how light behaves as it passes through a narrow slit or grating. The equation is expressed as: \(d \sin \theta = m \lambda\), where:
  • \(d\) represents the distance between the grating lines,
  • \(\theta\) is the angle at which a bright spot appears,
  • \(m\) denotes the order of the bright spot,
  • \(\lambda\) is the wavelength of the light.
This equation allows us to calculate either the angle of diffraction, the distance between grating lines, or any other unknown by manipulating the known variables. For example, if you know the wavelength and the order of the bright spot, you can rearrange the equation to find the grating spacing or angle.In practical applications, this equation enables us to precisely determine the structure of a diffraction pattern. This is essential for analyzing light spectra and understanding optical properties in various fields, including physics and engineering.
Light Spectrum
The **light spectrum** includes all possible wavelengths of electromagnetic radiation. Visible light is just a small portion of this spectrum, ranging from about 400 nm (violet) to 700 nm (red). Different colors are perceived because objects emit or reflect light at different wavelengths, with each color in the visible spectrum having its unique range:
  • Violet: 380-450 nm
  • Blue: 450-495 nm
  • Green: 495-570 nm
  • Yellow: 570-590 nm
  • Orange: 590-620 nm
  • Red: 620-750 nm
Understanding the light spectrum is crucial for applications like spectroscopy, where scientists study how light interacts with matter. By examining the spectrum of light, one can gain insights into the properties of different materials and identify substances based on their spectral lines.

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Most popular questions from this chapter

Laser light of wavelength 632.8 nm falls normally on a slit that is 0.0250 mm wide. The transmitted light is viewed on a distant screen where the intensity at the center of the central bright fringe is 8.50 W/m\(^2\). (a) Find the maximum number of totally dark fringes on the screen, assuming the screen is large enough to show them all. (b) At what angle does the dark fringe that is most distant from the center occur? (c) What is the maximum intensity of the bright fringe that occurs immediately before the dark fringe in part (b)? Approximate the angle at which this fringe occurs by assuming it is midway between the angles to the dark fringes on either side of it.

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.350 mm wide. The diffraction pattern is observed on a screen 3.00 m away. Define the width of a bright fringe as the distance between the minima on either side. (a) What is the width of the central bright fringe? (b) What is the width of the first bright fringe on either side of the central one?

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)m. (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

A diffraction grating has 650 slits>mm. What is the highest order that contains the entire visible spectrum? (The wavelength range of the visible spectrum is approximately 380-750 nm.)

(a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time \(t\), the electric field at a distant point \(P\) is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where \(E_0\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi = (2\pi d\) sin \(\theta)/\lambda\), \(\theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrangement), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\). (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship \(e^{iz} = cos z + i \space sin \space z\), where \(i = \sqrt{-1}\). In this expression, cos \(z\) is the \(real\) \(part\) of the complex number \(e^{iz}\), and sin \(z\) is its \(imaginary\) \(part\). Show that the electric field \(E_P(t)\) is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that \(e^Ae^B = e^{(A+B)}\) and \((e^A)^n = e^{nA}\), show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship \(e^{iz}\) = cos \(z\) + \(i\) sin \(z\), show that the (real) electric field at point \(P\) is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at \(P\). (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where \(I_0\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N\) = 2. It will help to recall that sin 2\(A\) = 2 sin \(A\) cos \(A\). Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (\(Hint\): Is I0 defined in the same way in both expressions?)
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