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Two different wavelengths of light are incident on a diffraction grating. One wavelength is \(600 . \mathrm{nm}\) and the other is unknown. If the 3 rd order of the unknown wavelength appears at the same position as the 2 nd order of the \(600 . \mathrm{nm}\) light, what is the value of the unknown wavelength?

Short Answer

Expert verified
Answer: The unknown wavelength of light is 400 nm.

Step by step solution

01

Understand the diffraction grating formula.

The diffraction grating formula relates the order of the pattern (m), the wavelength of light (位), the diffraction angle (胃), and the spacing between the grating lines (d). It is given as follows: m位 = dsin胃 In this problem, we are given that the 3rd order of the unknown wavelength appears at the same position as the 2nd order of the 600 nm light. We can use the diffraction grating formula to relate these orders and wavelengths.
02

Set up the equations for each wavelength.

We can first set up the equation for the known 600 nm light using the 2nd order and the grating formula: 2(600 nm) = d sin胃鈧 (1) Now we can set up a similar equation for the unknown wavelength using the 3rd order: 3 位鈧 = d sin胃鈧 (2) Since the 2nd order of the 600 nm light and the 3rd order of the unknown wavelength appear at the same position, we can say that 胃鈧 = 胃鈧
03

Divide the equations and solve for the unknown wavelength.

We can now divide equation (1) by equation (2) to eliminate the d and sin胃 terms, and solve for 位鈧. (2(600 nm)) / (3 位鈧) = (d sin胃鈧) / (d sin胃鈧) Since 胃鈧 = 胃鈧, this simplifies to: (2(600 nm)) / (3 位鈧) = 1 Now multiply both sides by 3位鈧 to isolate the unknown wavelength: 2(600 nm) = 3位鈧 Finally, divide both sides by 3 to find the value of 位鈧: 位鈧 = (2(600 nm)) / 3 位鈧 = 1200 nm / 3 位鈧 = 400 nm The unknown wavelength is 400 nm.

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Key Concepts

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

Diffraction Grating Formula
Understanding the diffraction grating formula is crucial when studying the properties of light as it interacts with a diffraction grating. A diffraction grating is made up of many equally spaced slits that diffract light into several beams. The formula itself is fairly simple in appearance but has a profound effect on how the different wavelengths of light are separated and analyzed.

The basic diffraction grating formula is expressed as:
\[ m\lambda = d\sin\theta \]
Here, \( m \) represents the order of the maxima, which essentially corresponds to how many 'humps' or bright spots you see on the pattern. The variable \( \lambda \) is the wavelength of the incident light, \( d \) is the spacing between grating lines, and \( \theta \) is the angle at which light is diffracted.

When you shine light through a diffraction grating, you can see these bright spots or lines at different angles, depending on the wavelength of the light. It鈥檚 these angles that give you information about the wavelength, as the formula shows how they're related mathematically.
Wavelength Calculation
To calculate the wavelength of light using a diffraction grating, one should first understand the relationship between the observed pattern and the light source's properties. Wavelength calculation is effectively done by rearranging the diffraction grating formula to solve for \( \lambda \).

Here's how you might calculate the wavelength when you have a known angle and diffraction order. Starting from the diffraction grating equation:
\[ m\lambda = d\sin\theta \]
We can rearrange this to find the wavelength, \( \lambda \), as follows:
\[ \lambda = \frac{{d\sin\theta}}{{m}} \]
It鈥檚 critical to understand that the diffraction grating creates a pattern that has multiple orders, with each order corresponding to a whole number multiple of wavelengths fitting into the path difference between adjacent slits. By observing these orders, you can use the formula to solve for the unknown wavelength when the angle and order are known, as demonstrated in the exercise provided.
Interference and Diffraction
Interference and diffraction are both fundamental concepts that describe how waves behave when they encounter obstacles or openings. While interference is the process where waves overlap and combine to form a new wave pattern, diffraction is specific to the bending and spreading of waves as they pass around corners or through slits.

In a diffraction grating problem, we generally observe constructive interference at specific angles where the waves from each slit reinforce each other, creating the bright fringes or maxima. The dark areas in between, known as minima, are the result of destructive interference, where waves cancel each other out. Both phenomena are governed by the principle of superposition, which states that when two or more waves overlap, the resultant wave is the sum of the individual waves.

Therefore, the interaction of light with a diffraction grating showcases both interference and diffraction. The specific pattern you see (bright and dark fringes) is a result of the constructive and destructive interference of light waves diffracted by the many slits in the grating. This interference pattern can be analyzed using the diffraction grating formula to extract information about the light's wavelength and other characteristics.

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Most popular questions from this chapter

A laser beam with wavelength \(633 \mathrm{nm}\) is split into two beams by a beam splitter. One beam goes to Mirror \(1,\) a distance \(L\) from the beam splitter, and returns to the beam splitter, while the other beam goes to Mirror \(2,\) a distance \(L+\Delta x\) from the beam splitter, and returns to the same beam splitter. The beams then recombine and go to a detector together. If \(L=1.00000 \mathrm{~m}\) and \(\Delta x=1.00 \mathrm{~mm},\) which best describes the kind of interference at the detector? (Hint: To doublecheck your answer, you may need to use a formula that was originally intended for combining two beams in a different geometry, but which still is applicable here.) a) purely constructive b) purely destructive c) mostly constructive d) mostly destructive e) neither constructive nor destructive

The thermal stability of a Michelson interferometer can be improved by submerging it in water. Consider an interferometer that is submerged in water, measuring light from a monochromatic source that is in air. If the movable mirror moves a distance of \(d=0.200 \mathrm{~mm},\) exactly \(N=800\) fringes move by the screen. What is the original wavelength (in air) of the monochromatic light?

Calculate and compare the angular resolutions of the Hubble Space Telescope (aperture diameter \(2.40 \mathrm{~m}\), wavelength \(450 . \mathrm{nm}\); illustrated in the text), the Keck Telescope (aperture diameter \(10.0 \mathrm{~m}\), wavelength \(450 . \mathrm{nm}\) ), and the Arecibo radio telescope (aperture diameter \(305 \mathrm{~m}\), wavelength \(0.210 \mathrm{~m}\) ). Assume that the resolution of each instrument is diffraction limited.

In Young's double-slit experiment, both slits were illuminated by a laser beam and the interference pattern was observed on a screen. If the viewing screen is moved farther from the slit, what happens to the interference pattern? a) The pattern gets brighter. b) The pattern gets brighter and closer together. c) The pattern gets less bright and farther apart. d) There is no change in the pattern. e) The pattern becomes unfocused. f) The pattern disappears.

A double slit is positioned in front of an incandescent light bulb. Will an interference pattern be produced?

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