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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

The intensity of light in the Fraunhofer diffraction pattern of a single slit is given by Eq. (36.5). Let \(\gamma=\beta / 2\). (a) Show that the equation for the values of \(\gamma\) at which \(I\) is a maximum is \(\tan \gamma=\gamma\). (b) Determine the two smallest positive values of \(\gamma\) that are solutions of this equation. (Hint: You can use a trial-and-error procedure. Guess a value of \(\gamma\) and adjust your guess to bring \(\tan \gamma\) closer to \(\gamma\). A graphical solution of the equation is very helpful in locating the solutions approximately, to get good initial guesses.) (c) What are the positive values of \(\gamma\) for the first, second, and third minima on one side of the central maximum? Are the \(\gamma\) values in part (b) precisely halfway between the \(\gamma\) values for adjacent minima? (d) If \(a=12 \lambda,\) what are the angles \(\theta\) (in degrees) that locate the first minimum, the first maximum beyond the central maximum, and the second minimum?

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

The Hubble Space Telescope has an aperture of \(2.4 \mathrm{~m}\) and focuses visible light \((380-750 \mathrm{nm})\). The Arecibo radio telescope in Puerto Rico is \(305 \mathrm{~m}(1000 \mathrm{ft})\) in diameter (it is built in a mountain valley) and focuses radio waves of wavelength \(75 \mathrm{~cm}\). (a) Under optimal viewing conditions, what is the smallest crater that each of these telescopes could resolve on our moon? (b) If the Hubble Space Telescope were to be converted to surveillance use, what is the highest orbit above the surface of the earth it could have and still be able to resolve the license plate (not the letters, just the plate) of a car on the ground? Assume optimal viewing conditions, so that the resolution is diffraction limited.

If a diffraction grating produces a third-order bright spot for red light (of wavelength \(700 \mathrm{nm}\) ) at \(65.0^{\circ}\) from the central maximum, at what angle will the second-order bright spot be for violet light (of wavelength \(400 \mathrm{nm}\) )?

An opaque barrier has an inner membrane and an outer membrane that slide past each other, as shown in Fig. \(\mathbf{P 3 6 . 6 5 .}\) Each membrane includes parallel slits of width \(a\) separated by a distance \(d\). A screen forms a circular arc subtending \(60^{\circ}\) at the fixed midpoint between the slits. A green \(532 \mathrm{nm}\) laser impinges on the slits from the left. The outer membrane moves upward with speed \(v\) while the inner membrane moves downward with the same speed, propelled by nanomotors. At time \(t=0,\) point \(P\) on the outer membrane is adjacent to point \(Q\) on the inner membrane so that the effective aperture width is zero. The aperture is fully closed again at \(t=3.00 \mathrm{~s}\). (a) At \(t=1.00 \mathrm{~s}\), there are 19 evenly spaced bright spots on the screen, each of approximately the same intensity. At the edges of the screen the first diffraction minimum and a two-slit interference maximum coincide. What is the slit distance \(d ?\) (Note: The screen does not encompass the entire diffraction pattern.) (b) What is the speed \(v ?\) (c) What is the maximum aperture width \(a ?\) (d) At a certain time, the outermost spots (the \(m=\pm 9\) spots) disappear. What is that time? (e) At \(t=1.50 \mathrm{~s}\) what is the intensity of the \(m=\pm 1\) spots in terms of the \(m=0\) central spot? (f) What are the angular positions of these spots?

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