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(a) What is the wavelength of light that is deviated in the first order through an angle of 13.5\(^\circ\) by a transmission grating having 5000 slits/cm? (b) What is the second-order deviation of this wavelength? Assume normal incidence.

Short Answer

Expert verified
(a) 468 nm, (b) 27.9°

Step by step solution

01

Understand the Problem

We need to find the wavelength of light that is deviated in the first order and the second-order deviation angle given a transmission grating with 5000 slits/cm. The angle for the first order is 13.5°.
02

Determine the Grating Equation

The grating equation is given by \( d\sin(\theta) = m\lambda \), where \( d \) is the distance between slits (grating spacing), \( \theta \) is the angle of deviation, \( m \) is the order of diffraction, and \( \lambda \) is the wavelength. For this problem, \( m = 1 \) for the first-order deviation.
03

Calculate Grating Spacing

The number of slits per cm is 5000. Hence, the grating spacing \( d \) is the inverse of this number. \( d = \frac{1}{5000 \text{ cm}^{-1}} = 2 \times 10^{-4} \text{ cm} = 2 \times 10^{-6} \text{ m}.\)
04

Solve for First Order Wavelength

Given \( \theta = 13.5^\circ \), convert it to radians for calculation: \( \theta = 13.5 \times \frac{\pi}{180} \). Use the equation \( d\sin(\theta) = m\lambda \). For \( m = 1 \), \( \lambda = d\sin(\theta) \). Computing gives \( \lambda = 2 \times 10^{-6} \sin(13.5^\circ) \).
05

Calculate for First Order Wavelength

Compute \( \lambda = 2 \times 10^{-6} \times \sin(13.5^\circ) \approx 4.68 \times 10^{-7} \text{ m} \) or 468 nm.
06

Calculate Second Order Deviation

Using the same formula, for \( m = 2 \), we rearrange to find \( \sin(\theta_2) = \frac{2 \lambda}{d} \). Substitute \( \lambda = 468\text{ nm} = 4.68 \times 10^{-7} \text{ m} \) to get \( \sin(\theta_2) = \frac{2 \times 4.68 \times 10^{-7}}{2 \times 10^{-6}} \approx 0.468 \).
07

Solve for Second Order Angle

Calculate \( \theta_2 \) by taking the inverse sine. \( \theta_2 = \arcsin(0.468) \approx 27.9^\circ \).

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Key Concepts

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

Wavelength Calculation
Wavelength calculation is a necessary step in various optical phenomena, including diffraction. In this exercise, we calculated the wavelength of light that undergoes diffraction through a grating. To find the wavelength (\(\lambda\)), we use the grating equation, but first, we need the distance between the slits, known as grating spacing (\(d\)). The spacing is the inverse of the number of slits per unit length. In this case, with 5000 slits per cm, \(d = \frac{1}{5000 \text{ cm}^{-1}} = 2 \times 10^{-6} \text{ m}\). The angle of deviation is provided in degrees, but it's useful for calculations to convert it to radians: \(\theta = 13.5 \times \frac{\pi}{180}\).Once you have these, the wavelength is calculated using \(\lambda = d\sin(\theta)\) for the first order (\(m = 1\)). Plug in the values:
  • \(d = 2 \times 10^{-6} \text{ m}\)
  • The sine of 13.5 degrees is used for \(\theta\)
This results in a wavelength of approximately 468 nm, which is a typical wavelength in the visible range.
Order of Diffraction
The order of diffraction is a significant concept in wave interference phenomena. It refers to the integer \(m\) in the grating equation that represents the sequence or hierarchy of the maxima.In this exercise, the "first order" implies that \(m = 1\), meaning the first level of constructive interference. For the given angle of 13.5 degrees, the calculation determines the wavelength at this initial order.The order of diffraction affects both the deviation angle and brightness of the diffracted waves. As \(m\) increases, the maxima become less intense, and diffraction patterns spread out. The second-order diffraction (\(m = 2\)) shows where this same wavelength of light deviates next. Calculations in this task were extended up to the second order to find an additional angle for this wavelength.
Grating Equation
The grating equation is the fundamental formula used to determine different properties of light as it passes through a diffraction grating. It is expressed as:\[ d\sin(\theta) = m\lambda \]where \(d\) is the grating spacing, \(\theta\) is the diffraction angle, \(m\) is the order of diffraction, and \(\lambda\) is the wavelength of light. This equation helps understand how different wavelengths are separated by a grating.For example, substituting into this equation helps calculate either the wavelength, the angle, or the order, depending on which values are known. The grating equation balances the optical characteristics and the physical arrangement of the grating, allowing for precise measurements. This equation shows why gratings are critical in applications like spectroscopy, where separating light into its component wavelengths is necessary.
Angle of Deviation
The angle of deviation is crucial in describing the behavior of light as it diffracts through a grating. It denotes the angle at which light emerges after passing through the grating slits, compared to the incident direction. In our exercise, the first order deviation occurred at an angle of 13.5 degrees. Calculating further, the second order angle was found to be approximately 27.9 degrees, showing a straightforward link between the order of diffraction and deviation angles. These angles are essential when designing optical instruments, as they dictate how separated the wavelengths become and how to align components to capture them. Notably, the angle of deviation depends not only on the order but also on the wavelength and grating spacing, as represented in the grating equation. This angle can further be fine-tuned by adjusting these parameters, enabling multiple experimental setups and analyses in optics.

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Most popular questions from this chapter

The maximum resolution of the eye depends on the diameter of the opening of the pupil (a diffraction effect) and the size of the retinal cells. The size of the retinal cells (about 5.0 \(\mu\)m in diameter) limits the size of an object at the near point (25 cm) of the eye to a height of about 50 \(\mu\)m. (To get a reasonable estimate without having to go through complicated calculations, we shall ignore the effect of the fluid in the eye.) (a) Given that the diameter of the human pupil is about 2.0 mm, does the Rayleigh criterion allow us to resolve a 50-\(\mu\)m- tall object at 25 cm from the eye with light of wavelength 550 nm? (b) According to the Rayleigh criterion, what is the shortest object we could resolve at the 25-cm near point with light of wavelength 550 nm? (c) What angle would the object in part (b) subtend at the eye? Express your answer in minutes (60 min = 1\(^\circ\)), and compare it with the experimental value of about 1 min. (d) Which effect is more important in limiting the resolution of our eyes: diffraction or the size of the retinal cells?

The wavelength range of the visible spectrum is approximately 380-750 nm. White light falls at normal incidence on a diffraction grating that has 350 slits/mm. Find the angular width of the visible spectrum in (a) the first order and (b) the third order. (\(Note\): An advantage of working in higher orders is the greater angular spread and better resolution. A disadvantage is the overlapping of different orders, as shown in Example 36.4.)

Monochromatic light with wavelength 490 nm passes through a circular aperture, and a diffraction pattern is observed on a screen that is 1.20 m from the aperture. If the distance on the screen between the first and second dark rings is 1.65 mm, what is the diameter of the aperture?

A single-slit diffraction pattern is formed by monochromatic electromagnetic radiation from a distant source passing through a slit 0.105 mm wide. At the point in the pattern 3.25\(^\circ\) from the center of the central maximum, the total phase difference between wavelets from the top and bottom of the slit is 56.0 rad. (a) What is the wavelength of the radiation? (b) What is the intensity at this point, if the intensity at the center of the central maximum is \(I_0\)?

Two satellites at an altitude of 1200 km are separated by 28 km. If they broadcast 3.6-cm microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

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