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Light of wavelength \(500 \mathrm{nm}\) is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at \(32.0^{\circ},\) (a) what is the number of rulings per centimeter for the grating? (b) Determine the total number of primary maxima that can be observed in this situation.

Short Answer

Expert verified
The number of rulings per centimeter of the grating is approximately 1667 rulings/cm. The total number of primary maxima that can be observed in this situation is 6.

Step by step solution

01

Finding the number of rulings per centimeter

We need to rearrange the diffraction grating equation to find \(d\): \(d = m\lambda / \sin \theta \). Let's convert units using the conversion factor \(1 \mathrm{cm} = 10^{-2} \mathrm{m}\) and \(1 \mathrm{nm} = 10^{-9} \mathrm{m}\). Now plug in the given values. The order \(m = 3\), wavelength \(\lambda = 500 \times 10^{-9}\) m and angle \(\theta = 32.0^{\circ}\). Calculate the value of \(d\) with these provided values.
02

Obtaining the number of rulings per centimeter

Once we get the value for \(d\) in meters, take its reciprocal and convert it to rulings per centimeter. The reciprocal \(1/d\) would yield the number of rulings per meter and multiplying it with 100 would give us the total number of rulings per centimeter.
03

Calculating the number of primary maxima

To obtain the total number of primary maxima observed, find the maximum value of \(m\), for which \(\sin \theta\) in the grating formula is less than or equal to 1. Rearrange the grating formula to yield \(m = d \sin \theta / \lambda\). We can now substitute \(d\) with the value obtained in step 1, \(\sin \theta\) with 1 (for maximum possible angle) and \(\lambda\) with 500 nm (converted back to meters). The maximum \(m\) will yield the total number of observable primary maxima.

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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 concept when studying waves, including light waves that undergo diffraction. It represents the distance between consecutive crests or troughs in a wave. In our exercise, we deal with a light wavelength of 500 nanometers (nm), which needs to be converted into meters for calculations. Since 1 nm equals 10-9 meters, our wavelength becomes \(500 \times 10^{-9}\) meters when converted.
Understanding wavelength is crucial in diffraction applications because it determines how light interacts with a diffraction grating, producing a diffraction pattern. The wavelength, in combination with the grating's properties, defines the angles where maximum and minimum points appear.
For practical purposes in diffraction gratings, a smaller wavelength relative to grating spacing will result in a more detailed diffraction pattern. In terms of measuring, noting wavelength's conversion from nanometers to meters is a frequent step in scientific calculations.
Diffraction Pattern
When light passes through a diffraction grating, it creates a diffraction pattern on a screen or detector. This pattern consists of various bright and dark spots, known as maxima and minima, respectively. The arrangement forms due to interference between the light waves as they encounter the grating.
The key to understanding the diffraction pattern lies in the principle of superposition, where overlapping waves result in increased or reduced intensity, leading to bright and dark regions. Bright spots, called primary maxima, occur at specific angles when the path difference between successive rulings (grooves) is an integral multiple of the wavelength.
To predict the angles at which these maxima occur, we use the diffraction grating formula: \(d \sin \theta = m \lambda\). Here, \(d\) is the distance between rulings, \(\theta\) is the angle of incidence or reflection, \(m\) is the order of the maximum, and \(\lambda\) is the wavelength. The pattern is crucial for understanding how waves behave when encountering barriers like slits or gratings.
Primary Maxima
Primary maxima are the principal bright spots in a diffraction pattern, occurring at angles where constructive interference takes place. They are distinguished by their brightness and spacing compared to other maxima, and a particular order, denoted by \(m\), indicates the sequence of these observable maxima.
To locate these primary maxima using a diffraction grating, we use the formula \(d \sin \theta = m \lambda\), which determines the angle \(\theta\) at which these maximum light intensities appear. The order \(m\) represents how many wavelengths fit into the path difference for the light reflected or transmitted through the grating.
In practical terms, finding the maximum value of \(m\), where \(\sin \theta\) is still valid (i.e., does not exceed 1), allows us to calculate the total number of primary maxima observable. This idea is central to analyzing the beam's spread and is vital in understanding limits to observation based on grating parameters.
Rulings per Centimeter
Rulings per centimeter describe the density of lines (or grooves) on a diffraction grating. This value influences the diffraction pattern's detail and resolution. More rulings imply a higher resolution, resulting in sharper and more distinct maxima.
In our exercise, finding the number of rulings per centimeter involves first calculating \(d\), the distance between these rulings, using the provided grating equation: \(d = \frac{m \lambda}{\sin \theta}\). Once \(d\) is determined, its reciprocal gives the number of rulings per meter— converting this to centimeters requires multiplying by 100, yielding the ruler density in rulings per centimeter.
Knowing the ruling density is crucial for predicting and engineering the level of detail in a diffraction pattern. Different applications, whether in spectroscopy or optical instruments, rely on the optimal ruling density to function accurately.

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Most popular questions from this chapter

You are vacationing in a Wonderland populated by friendly elves and a cannibalistic Cyclops that devours physics students. The elves and the Cyclops look precisely alike (everyone wears loose jeans and sweatshirts) except that each elf has two eyes, about \(10.0 \mathrm{cm}\) apart, and the Cyclops-you guessed it-has only one eye of about the same size as an elf's. The elves and the Cyclops are constantly at war with each other, so they rarely sleep and all have red eyes, predominantly reflecting light with a wavelength of \(660 \mathrm{nm} .\) From what maximum distance can you distinguish between a friendly elf and the predatory Cyclops? The air in Wonderland is always clear. Dilated with fear, your pupils have a diameter of \(7.00 \mathrm{mm}\)

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Helium-neon laser light \((\lambda=632.8 \mathrm{nm})\) is sent through a \(0.300-\mathrm{mm}\) -wide single slit. What is the width of the central maximum on a screen \(1.00 \mathrm{m}\) from the slit?

Two polarizing sheets are placed together with their transmission axes crossed so that no light is transmitted. A third sheet is inserted between them with its transmission axis at an angle of \(45.0^{\circ}\) with respect to each of the other axes. Find the fraction of incident unpolarized light intensity transmitted by the three-sheet combination. (Assume each polarizing sheet is ideal.)

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