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Chapter 13: Problem 50

When white light passes through a diffraction grating, what is the smallest value of \(m\) for which the visible spectrum of order \(m\) overlaps the next one, of order \(m+1 ?\)

Short Answer

Expert verified
The smallest value of \(m\) is 2.

Step by step solution

01

Understand the Principle of Diffraction Grating

When white light passes through a diffraction grating, it diffracts into its component colors (spectrum). For a diffraction grating, the condition for constructive interference is given by the equation: \( d \sin \theta = m \lambda \), where \(d\) is the distance between slits, \(\theta\) is the angle of diffraction, \(m\) is the order of the spectrum, and \(\lambda\) is the wavelength of light.
02

Define the Overlapping Condition

For the visible spectra of order \(m\) and \(m+1\) to overlap, the maximum wavelength of the spectrum from order \(m\) (red light) should coincide with the minimum wavelength from the order \(m+1\) (violet light). Let's denote the minimum visible wavelength as \(\lambda_{violet} = 400\,nm\) and the maximum as \(\lambda_{red} = 700\,nm\).
03

Establish the Overlap Equation

The overlap condition can be expressed as: \[ m \lambda_{red} = (m+1) \lambda_{violet} \] Substitute \(\lambda_{red} = 700\,nm\) and \(\lambda_{violet} = 400\,nm\) into the equation: \[ m \times 700 = (m+1) \times 400 \]
04

Solve for the Order \(m\)

Rearrange and solve the equation: \[ 700m = 400m + 400 \] \[ 300m = 400 \] \[ m = \frac{400}{300} = \frac{4}{3} \] Since \(m\) must be an integer, take the smallest integer greater than \(\frac{4}{3}\), which is \(m = 2\).

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Key Concepts

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

White Light Dispersion
White light is a mixture of all the colors in the visible spectrum. When it passes through a diffraction grating, this light spreads out into its component colors. Think of it like a prism effect, where white light splits into a rainbow.
A diffraction grating is specially designed to have many closely spaced slits or grooves. As light encounters these slits, different wavelengths (or colors) of light bend or 'diffract' at different angles.
  • Shorter wavelengths (like blue or violet) diffract at smaller angles.
  • Longer wavelengths (like red) spread out more, bending at larger angles.
Therefore, by using a diffraction grating, we can observe a full spectrum of colors spread out based on their wavelength.
Constructive Interference
In physics, when waves combine, they interfere. There are two main types: constructive and destructive interference. In the case of a diffraction grating, constructive interference is key to producing a visible spectrum.
Constructive interference occurs when waves align in such a way that their crests and troughs match up perfectly, producing a stronger combined wave.
  • For a diffraction grating, the condition for constructive interference is: \(d \sin \theta = m \lambda\).
  • Here, \(d\) is the spacing between the grating's slits, \(\theta\) is the angle of diffraction, \(m\) is the order of the spectrum, and \(\lambda\) is the wavelength of the light.
This equation ensures that the different wavelengths (colors) are separated out, producing visible bands.
Order of Spectrum
When discussing diffraction gratings, we often hear about 'orders' of the spectrum. This refers to the different sets or 'orders' of the separated spectrum that appear when light is diffracted. Each order represents a distinct pattern of colors.
Higher orders equate to more spread out colors, but this also leads to overlapping of colors from different orders.
  • Order \(m = 1\) is the first occurrence of the spectrum.
  • Order \(m = 2\) is the second, but occurs at a larger angle than the first.
  • Overlapping occurs when the red end of one order (higher wavelength) coincides with the violet end (lower wavelength) of the next, e.g., \(m\) and \(m+1\).
Understanding these orders helps in knowing how to set up and analyze experiments using diffraction gratings.

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Most popular questions from this chapter

When you take pictures with a camera, the distance between the lens and the film has to be adjusted, depending on the distance at which you want to focus. This is done by moving the lens. If you want to change your focus so that you can take a picture of something farther away, which way do you have to move the lens? Explain using ray diagrams.

There are two main types of telescopes, refracting (using a lens) and reflecting (using a mirror, as in figure i on p. 754 ). (Some telescopes use a mixture of the two types of elements: the light first encounters a large curved mirror, and then goes through an eyepiece that is a lens. To keep things simple, assume no eyepiece is used.) What implications would the colordependence of focal length have for the relative merits of the two types of telescopes? Describe the case where an image is formed of a white star. You may find it helpful to draw a ray diagram.

The beam of a laser passes through a diffraction grating, fans out, and illuminates a wall that is perpendicular to the original beam, lying at a distance of \(2.0 \mathrm{~m}\) from the grating. The beam is produced by a helium-neon laser, and has a wavelength of \(694.3 \mathrm{~nm}\). The grating has 2000 lines per centimeter. (a) What is the distance on the wall between the central maximum and the maxima immediately to its right and left? (b) How much does your answer change when you use the small-angle approximations \(\theta \approx \sin \theta \approx \tan \theta\) ?

Based on Snell's law, explain why rays of light passing through the edges of a converging lens are bent more than rays passing through parts closer to the center. It might seem like it should be the other way around, since the rays at the edge pass through less glass - - - shouldn't they be affected less? In your answer: \- Include a ray diagram showing a huge, full-page, close-up view of the relevant part of the lens. \- Make use of the fact that the front and back surfaces aren't always parallel; a lens in which the front and back surfaces are always parallel doesn't focus light at all, so if your explanation doesn't make use of this fact, your argument must be incorrect. \- Make sure your argument still works even if the rays don't come in parallel to the axis.

The figure shows a device for constructing a realistic optical illusion. Two mirrors of equal focal length are put against each other with their silvered surfaces facing inward. A small object placed in the bottom of the cavity will have its image projected in the air above. The way it works is that the top mirror produces a virtual image, and the bottom mirror then creates a real image of the virtual image. (a) Show that if the image is to be positioned as shown, at the mouth of the cavity, then the focal length of the mirrors is related to the dimension \(h\) via the equation $$\frac{1}{f}=\frac{1}{h}+\frac{1}{h+\left(\frac{1}{h}-\frac{1}{f}\right)^{-1}}$$ (b) Restate the equation in terms of a single variable \(x=h / f\), and show that there are two solutions for \(x\). Which solution is physically consistent with the assumptions of the calculation?
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