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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

Monochromatic light of wavelength 580 nm passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm\)90.0\(^\circ\), so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta\) = 45.0\(^\circ\) to the intensity at \(\theta\) = 0?

A thin slit illuminated by light of frequency \(f\) produces its first dark band at \(\pm\)38.2\(^\circ\) in air. When the entire apparatus (slit, screen, and space in between) is immersed in an unknown transparent liquid, the slit's first dark bands occur instead at \(\pm\)21.6\(^\circ\). Find the refractive index of the liquid.

Light of wavelength 585 nm falls on a slit 0.0666 mm wide. (a) On a very large and distant screen, how many \(totally\) dark fringes (indicating complete cancellation) will there be, including both sides of the central bright spot? Solve this problem \(without\) calculating all the angles! (\(Hint\): What is the largest that sin \theta can be? What does this tell you is the largest that \(m\) can be?) (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

A slit \(0.240 \mathrm{~mm}\) wide is illuminated by parallel light rays of wavelength \(540 \mathrm{nm} .\) The diffraction pattern is observed on a screen that is \(3.00 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(6.00 \times 10^{-6} \mathrm{W} / \mathrm{m}^{2}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the intensity at a point on the screen midway between the center of the central maximum and the first minimum?

Monochromatic light with wavelength 620 nm passes through a circular aperture with diameter 7.4 \(\mu\)m. The resulting diffraction pattern is observed on a screen that is 4.5 m from the aperture. What is the diameter of the Airy disk on the screen?

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