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Chapter 36: Problem 25

If a diffraction grating produces its third-order bright band at an angle of 78.4\(^\circ\) for light of wavelength 681 nm, find (a) the number of slits per centimeter for the grating and (b) the angular location of the first-order and second-order bright bands. (c) Will there be a fourth-order bright band? Explain.

Short Answer

Expert verified
13727 slits/cm; first-order at 23.4掳, second-order at 53.5掳; no fourth order.

Step by step solution

01

Understanding the Diffraction Formula

The diffraction grating formula relates the angles at which light of a certain wavelength will produce bright bands. It is given by \( d \sin \theta = m \lambda \), where \( d \) is the distance between slits, \( \theta \) is the angle of deflection, \( m \) is the order of the bright band, and \( \lambda \) is the wavelength of light.
02

Calculate Slit Separation for Third-Order Band

For the third-order band, \( m = 3 \). Given \( \theta = 78.4^{\circ} \) and \( \lambda = 681 \text{ nm} = 681 \times 10^{-9} \text{ m} \), we can rearrange the formula to find \( d \): \( d = \frac{3 \times 681 \times 10^{-9}}{\sin(78.4^{\circ})} \). Calculate \( \sin(78.4^{\circ}) \), then solve for \( d \).
03

Convert Slit Separation to Slits per Centimeter

We have the slit separation \( d \) in meters. To find the number of slits per centimeter, first convert \( d \) to centimeters and then take the reciprocal: \( ext{slits per cm} = \frac{1}{d\text{ in cm}} \).
04

First-Order Bright Band Calculation

For the first-order bright band, use \( m = 1 \). Solve the equation \( d \sin \theta = 1 \times 681 \times 10^{-9} \) for \( \theta \), using the determined \( d \) value from the previous steps: \( \sin \theta = \frac{681 \times 10^{-9}}{d} \) and solve for \( \theta \).
05

Second-Order Bright Band Calculation

For the second-order bright band, \( m = 2 \). Again, use the equation \( d \sin \theta = 2 \times 681 \times 10^{-9} \) to solve for \( \theta \) using the same \( d \): \( \sin \theta = \frac{2 \times 681 \times 10^{-9}}{d} \) and solve for \( \theta \).
06

Check for Fourth-Order Bright Band

For a fourth-order bright band, set \( m = 4 \). Use the equation \( d \sin \theta = 4 \times 681 \times 10^{-9} \). Since \( \sin \theta \leq 1 \), there will be no solution if \( \frac{4 \times 681 \times 10^{-9}}{d} > 1 \). Calculate this to confirm if the fourth-order band exists.

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Key Concepts

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

Diffraction Formula
The diffraction formula is a key concept in understanding how light interacts with a diffraction grating, which is a series of closely spaced slits. When light passes through these slits, it bends or diffracts, and can produce bright and dark bands on a screen. This phenomenon is described by the equation \( d \sin \theta = m \lambda \). Here, \( d \) stands for the distance between the slits 鈥� called the slit separation. \( \theta \) is the angle at which a bright band occurs, \( m \) represents the order number of the bright band (like first-order, second-order, etc.), and \( \lambda \) is the wavelength of the light you are using.
  • \( d \): Distance between slits
  • \( \theta \): Angle of diffraction
  • \( m \): Order of the bright band
  • \( \lambda \): Wavelength of light
The formula tells us that light of different wavelengths will diffract at different angles, creating distinct patterns of light and dark bands.
Wavelength of Light
Wavelength, denoted by \( \lambda \), is a fundamental attribute of light that significantly influences how it interacts with objects such as diffraction gratings. Essentially, the wavelength is the distance between consecutive points of a wave, like peaks or troughs. In the context of visible light, wavelengths are typically measured in nanometers (nm). For our exercise, the wavelength is 681 nm.This characteristic determines not only the color of the visible light but also how it bends when passing through or around obstacles. In a diffraction grating, the wavelength helps decide the angle at which the bright bands appear. Longer wavelengths bend less sharply than shorter ones鈥攎eaning that red light (longer wavelength) diffracts less than blue light (shorter wavelength).
Bright Band Orders
In diffraction, bright bands are the results of light waves meeting in such a way that their crests coincide, a process known as constructive interference. These bright spots occur at specific angles, and each is attributed an order number, \( m \).
  • First-order band: \( m = 1 \)
  • Second-order band: \( m = 2 \)
  • Third-order band: \( m = 3 \), and so on.
The order number indicates how many wavelengths fit into the path difference between adjacent slits as light travels to the screen. Higher orders appear at larger angles. In our problem, we calculate the angles for the first, second, and third orders. More importantly, for orders higher than three, the calculation may show that these cannot exist if the sine function exceeds 1, as physical constraints limit \( \sin \theta \leq 1 \). This means not all orders will always appear depending on the combination of the diffraction grating's properties and the light's wavelength.
Slits Per Centimeter
A crucial detail when working with diffraction gratings is knowing how many slits contribute to the diffraction process, usually expressed as slits per centimeter. This measure provides insight into the grating's density, with more slits per centimeter resulting in greater diffraction and more detailed separation of light.To find slits per centimeter, first get the slit separation \( d \) using the diffraction formula, then convert \( d \) into centimeters. The number of slits per centimeter is simply the reciprocal of this value. Larger values here mean more slits packed into a small space, enhancing the grating's ability to produce clear, distinct interference patterns. This concept allows us to understand and predict the angles at which light's bright bands will form for various orders.

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Most popular questions from this chapter

An interference pattern is produced by light of wavelength 580 nm from a distant source incident on two identical parallel slits separated by a distance (between centers) of 0.530 mm. (a) If the slits are very narrow, what would be the angular positions of the first-order and second-order, two-slit interference maxima? (b) Let the slits have width 0.320 mm. In terms of the intensity \(I_0\) at the center of the central maximum, what is the intensity at each of the angular positions in part (a)?

Light of wavelength 633 nm from a distant source is incident on a slit 0.750 mm wide, and the resulting diffraction pattern is observed on a screen 3.50 m away. What is the distance between the two dark fringes on either side of the central bright fringe?

(a) Consider an arrangement of \(N\) slits with a distance \(d\) between adjacent slits. The slits emit coherently and in phase at wavelength \(\lambda\). Show that at a time \(t\), the electric field at a distant point \(P\) is $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ where \(E_0\) is the amplitude at \(P\) of the electric field due to an individual slit, \(\phi = (2\pi d\) sin \(\theta)/\lambda\), \(\theta\) is the angle of the rays reaching \(P\) (as measured from the perpendicular bisector of the slit arrangement), and \(R\) is the distance from \(P\) to the most distant slit. In this problem, assume that \(R\) is much larger than \(d\). (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship \(e^{iz} = cos z + i \space sin \space z\), where \(i = \sqrt{-1}\). In this expression, cos \(z\) is the \(real\) \(part\) of the complex number \(e^{iz}\), and sin \(z\) is its \(imaginary\) \(part\). Show that the electric field \(E_P(t)\) is equal to the real part of the complex quantity $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that \(e^Ae^B = e^{(A+B)}\) and \((e^A)^n = e^{nA}\), show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship \(e^{iz}\) = cos \(z\) + \(i\) sin \(z\), show that the (real) electric field at point \(P\) is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ The quantity in the first square brackets in this expression is the amplitude of the electric field at \(P\). (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle \(\theta\) is $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ where \(I_0\) is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case \(N\) = 2. It will help to recall that sin 2\(A\) = 2 sin \(A\) cos \(A\). Explain why your result differs from Eq. (35.10), the expression for the intensity in two-source interference, by a factor of 4. (\(Hint\): Is I0 defined in the same way in both expressions?)

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?

On December 26, 2004, a violent earthquake of magnitude 9.1 occurred off the coast of Sumatra. This quake triggered a huge tsunami (similar to a tidal wave) that killed more than 150,000 people. Scientists observing the wave on the open ocean measured the time between crests to be 1.0 h and the speed of the wave to be 800 km/h. Computer models of the evolution of this enormous wave showed that it bent around the continents and spread to all the oceans of the earth. When the wave reached the gaps between continents, it diffracted between them as through a slit. (a) What was the wavelength of this tsunami? (b) The distance between the southern tip of Africa and northern Antarctica is about 4500 km, while the distance between the southern end of Australia and Antarctica is about 3700 km. As an approximation, we can model this wave's behavior by using Fraunhofer diffraction. Find the smallest angle away from the central maximum for which the waves would cancel after going through each of these continental gaps.
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