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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 \(/ \mathrm{cm} ?\) (b) What is the second-order deviation of this wavelength? Assume normal incidence.

Short Answer

Expert verified
The wavelength of the light deviated in the first order is approximately 478 nm. This wavelength is deviated by an angle of approximately 28.43° in the second order.

Step by step solution

01

Convert number of slits to meter

5000 slits / cm equates to \(5 \times 10^{5}\) slits / m because 1m equals 100 cm. This translates to a slit spacing, \(d\), of \(1 / 5 \times 10^{5} = 2 \times 10^{-6}\) m.
02

Calculate Wave Length

Solving for \(\lambda\) in the formula, we get \(\lambda = \frac {d \cdot sin( \theta )}{m}\). Given that \(m = 1\) for first-order deviation, \( \theta = 13.5^{\circ} = \frac {13.5 \cdot \pi}{180^{\circ}} \) in radians. Therefore, \(\lambda = \frac {2 \times 10^{-6} sin( \frac {13.5 \cdot \pi}{180})}{1} = 4.78 \times 10^{-7}\) m, or approximately 478 nm (in nanometers).
03

Calculate Second-Order Deviation

For the second order deviation, we'll have to calculate the angle that the wavelength \( \lambda = 4.78 \times 10^{-7} m \) was deviated. Therefore, we rearrange the formula to: \( \theta = arcsin( \frac {m \cdot \lambda}{d}) \). Plugging in the values \( m = 2 \), \( \lambda = 4.78 \times 10^{-7} m \) and \( d = 2 \times 10^{-6} m \), we get \( \theta = arcsin( \frac {2 \cdot 4.78 \times 10^{-7}}{2 \times 10^{-6}})\). This translates to \( \theta = arcsin(0.478) = 28.43^{\circ}\).

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Key Concepts

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

Wavelength Calculation
When light waves pass through a diffraction grating, they spread out and create a pattern based on their wavelengths. Determining the wavelength involves understanding the relationship between the grating's properties and the angle at which light is diffracted. Let's explore how to calculate the wavelength of light using a diffraction grating.
The key formula to remember is:
  • \( \lambda = \frac{d \cdot \sin(\theta)}{m} \)

Here, \( \lambda \) represents the wavelength of light. The variable \( d \) is the distance between adjacent slits in the grating, \( \theta \) is the diffraction angle, and \( m \) is the order of diffraction, which indicates how many times the pattern repeats.
To find the wavelength:- Convert the number of slits per cm into slit distance in meters. For example, 5000 slits per cm results in a spacing of \( 2 \times 10^{-6} \) m.- Input the first-order diffraction angle into the formula. Given an angle of \( 13.5^{\circ} \), convert this into radians for calculation.- Apply the values into the formula to compute the wavelength, resulting in \( \lambda = 478 \) nm.
Understanding these basics will help you grasp how different wavelengths are distinguished by a diffraction grating.
Order of Diffraction
Diffraction gratings produce multiple sequences or orders of light patterns as the light waves are dispersed. Each order represents a distinct way the light spectrum is spread out by the grating. Grasping the concept of diffraction orders is important to fully understand how diffraction gratings function.
A diffraction order is represented by an integer \( m \), which indicates the sequence number of the pattern. The first-order diffraction \( m = 1 \) indicates the primary set of fringes closest to the original direction of light. Subsequent orders, like second-order \( m = 2 \), show additional sets of fringes that are further separated.
The position and clarity of these fringes depend on the wavelength of the light, the angle of incidence, and the grating's properties. Since
  • \( \lambda = \frac{d \cdot \sin(\theta)}{m} \)
Changing \( m \) in calculations shifts the angle \( \theta \), thus affecting the visibility and position of patterns.
Higher orders may become convoluted with overlapping spectra or can be more spread out. Exploring different orders reveals intricate light behaviors through diffraction, helping you appreciate the full spectrum of results a grating can generate.
Angle of Deviation
The angle of deviation in diffraction grating analyzes how far a light wave moves from its initial path due to diffraction. This concept is crucial for understanding how light bends and alters course as it passes through the slits of the grating.
The angle of deviation, \( \theta \), can be calculated using the rearranged form of the diffraction formula:
  • \( \theta = \arcsin\left(\frac{m \cdot \lambda}{d}\right) \)
Here, \( \theta \) is the angle of deviation, \( m \) is the diffraction order, \( \lambda \) is the wavelength, and \( d \) is the spacing between the grating slits.
For example, in second-order deviation calculations where \( m = 2\), and for a known wavelength of \( 478 \) nm:
  • First convert the wavelength into meters, resulting in \( 4.78 \times 10^{-7} \) m.
  • Plug the values into the formula to find \( \theta = 28.43^{\circ} \).
Understanding these angles can also assist in setting up experiments and analyzing light dispersion accurately. Such practical applications include separating different colors in a beam of light to study properties of materials or to produce clear spectral images.
Realizing how this angle shifts with varying wavelengths and orders will deepen your comprehension of light behavior through diffraction.

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Most popular questions from this chapter

A slit \(0.360 \mathrm{~mm}\) wide is illuminated by parallel rays of light that have a wavelength of \(540 \mathrm{nm}\). The diffraction pattern is observed on a screen that is \(1.20 \mathrm{~m}\) from the slit. The intensity at the center of the central maximum \(\left(\theta=0^{\circ}\right)\) is \(I_{0}\). (a) What is the distance on the screen from the center of the central maximum to the first minimum? (b) What is the distance on the screen from the center of the central maximum to the point where the intensity has fallen to \(I_{0} / 2 ?\)

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

X rays of wavelength \(0.0850 \mathrm{nm}\) are scattered from the atoms of a crystal. The second-order maximum in the Bragg reflection occurs when the angle \(\theta\) in Fig. 36.22 is \(21.5^{\circ} .\) What is the spacing between adjacent atomic planes in the crystal?

Monochromatic light of wavelength \(580 \mathrm{nm}\) passes through a single slit and the diffraction pattern is observed on a screen. Both the source and screen are far enough from the slit for Fraunhofer diffraction to apply. (a) If the first diffraction minima are at \(\pm 90.0^{\circ},\) so the central maximum completely fills the screen, what is the width of the slit? (b) For the width of the slit as calculated in part (a), what is the ratio of the intensity at \(\theta=45.0^{\circ}\) to the intensity at \(\theta=0 ?\)

Why is visible light, which has much longer wavelengths than x rays do, used for Bragg reflection experiments on colloidal crystals? (a) The microspheres are suspended in a liquid, and it is more difficult for x rays to penetrate liquid than it is for visible light. (b) The irregular spacing of the microspheres allows the longer-wavelength visible light to produce more destructive interference than can x rays. (c) The microspheres are much larger than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be much longer than the size of the individual scatterers. (d) The microspheres are spaced more widely than atoms in a crystalline solid, and in order to get interference maxima at reasonably large angles, the wavelength must be comparable to the spacing between scattering planes.

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