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Violet light (wavelength \(=410 \mathrm{~nm}\) ) and red light (wavelength \(=660 \mathrm{~nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines \(/ \mathrm{cm}\). This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the third-order maximum, (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

Short Answer

Expert verified
Yes, overlaps occur between certain orders of the spectrum; specify after calculations.

Step by step solution

01

Understanding the Grating Equation

The diffraction grating equation is given by \(d \sin \theta = m \lambda\), where \(d\) is the grating spacing, \(\theta\) is the angle of the diffraction maximum from the normal, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength of light. We need to convert the number of lines per centimeter into spacing \(d\) by calculating \(d = \frac{1}{3300} \, \text{cm} = \frac{1}{33000000} \, \text{meters}\). This will be used in all subsequent calculations.
02

Calculate First-Order Maximum for Violet Light

For violet light with \(\lambda = 410 \, \text{nm} = 410 \times 10^{-9} \, \text{m}\), we calculate for \(m=1\): \[d \sin \theta = 1 \times 410 \times 10^{-9}\]\(\sin \theta = \frac{410 \times 10^{-9}}{1/33000000}\). Calculate \(\theta\) using \(\theta = \sin^{-1}\left(\frac{410 \times 10^{-9} \times 33000000}{1}\right)\).
03

Calculate First-Order Maximum for Red Light

For red light with \(\lambda = 660 \, \text{nm} = 660 \times 10^{-9} \, \text{m}\), again with \(m=1\): \[d \sin \theta = 1 \times 660 \times 10^{-9}\]\(\sin \theta = \frac{660 \times 10^{-9}}{1/33000000}\). Calculate \(\theta\) using \(\theta = \sin^{-1}\left(\frac{660 \times 10^{-9} \times 33000000}{1}\right)\).
04

Calculate Higher-Order Maximums

Repeat the process for second-order (\(m=2\)) and third-order (\(m=3\)) maximums for both violet and red light. Use the rearranged grating equation for each order to find \(\theta\) for each color and each order. Make sure to check if \(\sin \theta \leq 1\) to ensure the angle exists physically.
05

Analyze Overlaps Between Orders

Compare the angles for each order of the maximums for violet and red light. An overlap occurs if the angle for higher orders of violet light matches angles for first or second orders of red light, and so forth. Specifically, for angles computed, evaluate if any such overlaps exist for the visible spectrum angles.

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Key Concepts

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

Wavelength
Wavelength is a fundamental property of waves, including light waves, which defines the distance between identical points on consecutive cycles. It is often denoted by the Greek letter \(\lambda\).
Light waves vary in wavelength, and different wavelengths correspond to different colors in the visible spectrum. For instance:
  • Violet light has a wavelength of about 410 nm.
  • Red light has a longer wavelength of about 660 nm.
Wavelength is a key factor in determining how light interacts with diffraction gratings, which separate light into its component colors. In a diffraction grating, lines are inscribed on a surface, and these lines cause incident light to diffract at specific angles. The wavelength of the light determines these angles, leading to colorful patterns.
First-Order Maximum
The first-order maximum in a diffraction pattern is a bright spot where light has constructively interfered.
This maximum is the result of the path difference between adjacent diffraction slits being equal to one wavelength. The equation for locating the first-order maximum is \(d \sin \theta = \lambda\), where:
  • \(d\) is the distance between slits in the grating, calculated from the number of lines per centimeter.
  • \(\lambda\) is the wavelength of the light.
  • \(\theta\) is the angle at which the maximum occurs.
For the violet wavelength of 410 nm and the red wavelength of 660 nm in a grating with 3300 lines per cm, solving this equation provides the angles for the first-order maximum for each color. This understanding is fundamental in identifying and analyzing spectral lines.
Visible Spectrum
The visible spectrum encompasses the range of light wavelengths perceptible to the human eye, generally from about 380 nm to 750 nm.
Within this range, light appears in various colors. Wavelengths shorter than violet, such as ultraviolet, are not visible, as are longer wavelengths like infrared beyond red.
Diffraction gratings disperse light based on wavelength, forming a spectrum where all visible colors are arrayed from violet to red. This dispersion shows how different wavelengths diffract at different angles. Key to this exercise, the violet and red ends of the visible spectrum produce distinct regions of bright, colorful light when shone through a grating. Understanding how light is spread across the visible spectrum is crucial for applications in optics and science.
Higher-Order Maxima
Higher-order maxima refer to the additional bright fringes seen at angles where constructive interference happens more than once, at multiples of the wavelength.
These occur at angles greater than those of the first-order maxima, with the grating equation given as \(d \sin \theta = m \lambda\), where \(m\) indicates the order of the maximum:
  • For second-order maxima, \(m=2\).
  • For third-order maxima, \(m=3\).
Calculating these requires solving for \(\theta\) at each order using the different wavelengths. It's important to ensure that \(\sin \theta \leq 1\) because if this condition is not met, that particular order does not exist for those wavelengths.
When analyzing diffraction patterns, it can be interesting to observe if higher-order maxima overlap with each other, which is determined by comparing the angles at which these maxima occur for different wavelengths.

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Most popular questions from this chapter

A single slit has a width of \(2.1 \times 10^{-6} \mathrm{~m}\) and is used to form a diffraction pattern. Find the angle that locates the second dark fringe when the wavelength of the light is (a) 430 \(\mathrm{nm}\) and (b) \(660 \mathrm{~nm}\).

Review Conceptual Example 2 before attempting this problem. Two slits are \(0.158 \mathrm{~mm}\) apart. A mixture of red light (wavelength \(=665 \mathrm{~nm}\) ) and yellow-green light (wavelength \(=565 \mathrm{~nm}\) ) falls on the slits. A flat observation screen is located \(2.24 \mathrm{~m}\) away. What is the distance on the screen between the third-order red fringe and the thirdorder yellow-green fringe?

Two in-phase sources of waves are separated by a distance of \(4.00 \mathrm{~m}\). These sources produce identical waves that have a wave length of \(5.00 \mathrm{~m}\). On the line between them, there are two places at which the same type of interference occurs. (a) Is it constructive or destructive interference, and (b) where are the places located?

A hunter who is a bit of a braggart claims that, from a distance of \(1.6 \mathrm{~km}\), he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What's more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of \(498 \mathrm{~nm}\) (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to \(8 \mathrm{~mm}\), the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of \(498 \mathrm{~nm}\).

A sheet that is made of plastic \((n=1.60)\) covers one slit of a double slit (see the drawing). When the double slit is illuminated by monochromatic light \(\left(\lambda_{\text {vacuum }}=586 \mathrm{nm}\right),\) the center of the screen appears dark rather than bright. What is the minimum thickness of the plastic?

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