/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 65 What is the longest wavelength t... [FREE SOLUTION] | 91Ó°ÊÓ

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What is the longest wavelength that can be observed in the third order for a transmission grating having 6500 slits \(/ \mathrm{cm} ?\) Assume normal incidence.

Short Answer

Expert verified
The longest wavelength that can be observed is 513 nm.

Step by step solution

01

Understand the Grating Equation

The basic formula for diffraction grating is given by: \( d \sin \theta = m \lambda \). Here, \(d\) is the distance between adjacent slits, \(\theta\) is the angle of diffraction, \(m\) is the order of diffraction, and \(\lambda\) is the wavelength. For normal incidence (where \(\theta = 90^\circ\)), the formula simplifies to \(d = \frac{m\lambda}{\sin \theta}\).
02

Determine the Grating Constant

The grating constant \(d\) is the distance between each slit. Since there are 6500 slits per centimeter, \(d\) in centimeters is \( \frac{1}{6500} \). Converting to meters for convenience, this becomes \(d = \frac{1}{6500} \times 10^{-2} \ \text{m} \).
03

Set Up Equation for Maximum Wavelength

We are asked to find the longest wavelength for the third order, i.e., \(m = 3\). At maximum wavelength, the angle \(\theta\) is maximum, signifying \(\sin \theta = 1\). Therefore, substitute \(\sin \theta = 1\) into the simplified grating equation: \(d = \frac{3 \lambda}{1}\).
04

Solve for Maximum Wavelength

Now solve for \(\lambda\):\[ \lambda = \frac{d}{3} = \frac{\frac{1}{6500} \times 10^{-2}}{3} = \frac{10^{-2}}{6500 \times 3} \]Calculate this result:\[ \lambda = \frac{10^{-2}}{19500} = 5.13 \times 10^{-7} \ \text{m} \]
05

Convert Wavelength to Nanometers

Convert \(\lambda\) from meters to nanometers by multiplying by \(10^9\) (since 1 nm = \(10^{-9}\) m):\[ \lambda = 5.13 \times 10^{-7} \times 10^9 = 513 \ \text{nm} \].

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Key Concepts

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

Wavelength Calculation
To determine the longest wavelength observed through a diffraction grating, we utilize the grating equation: \[ d \sin \theta = m \lambda \]where:- \(d\) represents the distance between slits, known as the grating constant,- \(\theta\) is the angle of diffraction,- \(m\) signifies the order of diffraction,- \(\lambda\) denotes the wavelength.
In our example, the wavelength is calculated by rearranging the equation to solve for \(\lambda\): \[ \lambda = \frac{d \sin \theta}{m} \]For the longest wavelength, we assume \(\sin \theta = 1\), resulting in:\[ \lambda = \frac{d}{m} \]The calculated wavelength should then be converted to appropriate units like nanometers, by multiplying by \(10^9\). This process ensures accuracy in finding the longest wavelength for the specified diffraction order.
Order of Diffraction
The term 'order of diffraction' refers to the number of wavelengths by which paths differ as they spread out from the grating. Indicated as \(m\) in diffraction equations, each integer value represents a distinct diffraction order. It is essentially a count of the constructive interference patterns observable when light diffracts through a grating.
Higher orders (larger \(m\)) correspond to greater angles, which are formed at larger wavelengths or when using gratings with fewer slits per unit area. For analyzing the longest wavelength in such scenarios, especially in multiple orders of diffraction like the third order \(m = 3\), it’s crucial to correctly apply the grating equation.
Grating Constant
The grating constant \(d\) is the separation between two adjacent slits in a diffraction grating. It is pivotal in the grating equation, determining how light of different wavelengths will diffract. The value of \(d\) is calculated from the number of slits per unit length.
In the given problem, there are 6500 slits per centimeter. This converts to a grating constant in meters as: \[ d = \frac{1}{6500} \times 10^{-2} \mid \text{m} \] The conversion is essential because it aligns with standard units used in scientific calculations. A smaller \(d\) means more slits per unit length, which typically allows for a more defined diffraction pattern.
Angle of Diffraction
The angle of diffraction \(\theta\) defines how far a beam of light deviates as it passes through a grating. It is influenced by the wavelength of the light \(\lambda\), the order of diffraction \(m\), and the grating constant \(d\).
In normal incidence situations, like our exercise, light hits the grating directly, optimizing the diffraction spread. The angle varies with orders of diffraction, such that for the longest wavelength experiment, it's assumed that \(\theta\) reaches a maximum when \(\sin \theta = 1\). This implies that good knowledge of \(\theta\)'s relationships helps in practical settings, like designing optical instruments or understanding the course of light in diffraction phenomena.

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Most popular questions from this chapter

Observing Jupiter. You are asked to design a space telescope for earth orbit. When Jupiter is \(5.93 \times 10^{8} \mathrm{km}\) away (its closest approach to the earth), the telescope is to resolve, by Rayleigh's criterion, features on Jupiter that are 250 \(\mathrm{km}\) apart. What minimum-diameter mirror is required? Assume a wavelength of 500 \(\mathrm{nm}\) .

Parallel rays of light with wavelength 620 \(\mathrm{nm}\) pass through a slit covering a lens with a focal length of 40.0 \(\mathrm{cm}\) . The diffraction pattern is observed in the focal plane of the lens, and the distance from the center of the central maximum to the first minimum is \(36.5 \mathrm{cm} .\) What is the width of the slit? (Note: The angle that locates the first minimum is not small.)

Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 \(\mathrm{m}\) away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 \(\mathrm{mm}\) . Calculate the wavelength of the light.

Light of wavelength 585 \(\mathrm{nm}\) falls on a slit 0.0666 \(\mathrm{mm}\) wide. (a) On a very large 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 calclating all the anglest (Hint: What is the largest that sin \(\theta\) can be? What does this tell you is the largest that \(m\) can be 7 (b) At what angle will the dark fringe that is most distant from the central bright fringe occur?

A series of parallel linear water wave fronts are traveling directly toward the shore at 15.0 \(\mathrm{cm} / \mathrm{s}\) on an otherwise placid lake. A long concrete barrier that runs parallel to the shore at a distance of 3.20 \(\mathrm{m}\) away has a hole in it. You count the wave crests and observe that 75.0 of them pass by each minute, and you also observe that no waves reach the shore at \(\pm 61.3 \mathrm{cm}\) from the point directly opposite the hole, but waves do reach the shore everywhere within this distance. (a) How wide is the hole in the barrier? (b) At what other angles do you find no waves hitting the shore?

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