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A diffraction grating \(20.0 \mathrm{~mm}\) wide has 6000 rulings. Light of wavelength \(589 \mathrm{nm}\) is incident perpendicularly on the grating. What are the (a) largest, (b) second largest, and (c) third largest values of \(\theta\) at which maxima appear on a distant viewing screen?

Short Answer

Expert verified
(a) \( 62.3^\circ \), (b) \( 45.0^\circ \), (c) \( 32.0^\circ \).

Step by step solution

01

Calculate the grating spacing (d)

The diffraction grating has 6000 rulings in a width of 20.0 mm. The spacing between the rulings, often denoted as \( d \), is therefore \( d = \frac{20.0 \, \text{mm}}{6000} = \frac{20.0 \, \times \ 10^{-3} \, \text{m}}{6000} \). Calculating this gives \( d = 3.33 \, \times \ 10^{-6} \, \text{m} = 3333 \, \text{nm} \).
02

Use the diffraction formula

Diffraction maxima occur when \( d \sin \theta = m \lambda \), where \( m \) is the order of the maximum, \( d \) is the grating spacing calculated above, and \( \lambda \) is the wavelength \( 589 \, \text{nm} \).
03

Find the maximum order (m) for which diffraction maxima appear

To find the maximum \( m \), use the condition \( \sin \theta \leq 1 \). Thus, \( m \leq \frac{d}{\lambda } \). Substituting the values, \( m \leq \frac{3333}{589} \approx 5.66 \). Therefore, the highest integer value is \( m = 5 \).
04

Calculate the largest angle for maxima (\( m=5 \))

Using the formula \( \sin(\theta) = \frac{m \lambda}{d} \) for \( m = 5 \), we find \( \sin(\theta_5) = \frac{5 \times 589}{3333} \approx 0.884 \). Then, \( \theta_5 = \arcsin(0.884) \approx 62.3^\circ \).
05

Calculate the second largest angle for maxima (\( m=4 \))

For \( m = 4 \), \( \sin(\theta_4) = \frac{4 \times 589}{3333} \approx 0.707 \). Thus, \( \theta_4 = \arcsin(0.707) \approx 45.0^\circ \).
06

Calculate the third largest angle for maxima (\( m=3 \))

For \( m = 3 \), \( \sin(\theta_3) = \frac{3 \times 589}{3333} \approx 0.530 \). Hence, \( \theta_3 = \arcsin(0.530) \approx 32.0^\circ \).

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Key Concepts

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

Grating Spacing
Grating spacing is a crucial concept when dealing with diffraction gratings. It refers to the separation between adjacent rulings on a grating and is often denoted by the symbol \( d \). For the given problem, a diffraction grating has a width of 20.0 mm and features 6000 rulings. The formula to calculate grating spacing is:
  • \( d = \frac{W}{N} \)
Here, \( W \) is the total width of the grating, and \( N \) is the number of rulings. Substituting the given values, \( d = \frac{20.0 \, \text{mm}}{6000} = 3.33 \, \times \, 10^{-6} \, \text{m} \). This result is also expressed as \( 3333 \, \text{nm} \) for consistency with wavelength units. Having accurate grating spacing is essential for calculations involving diffraction patterns.
Diffraction Maxima
In optics, diffraction maxima occur when waves overlap constructively, leading to bright bands on a viewing screen. This phenomenon is mathematically expressed by the formula:
  • \( d \sin \theta = m \lambda \)
Here, \( d \) is the grating spacing, \( \lambda \) is the wavelength of the light, and \( m \) denotes the order of the maximal intensity points, starting with \( m = 0, 1, 2, \) and so on. The condition \( \sin \theta \leq 1 \) restricts the maximum order \( m \) and thus determines the possible angles \( \theta \). In the example provided, the integer values of \( m \) are found by dividing \( d \) by \( \lambda \) and rounding down to the closest integer, which in this case allows up to \( m = 5 \). This allows us to compute the angles \( \theta \) for each consecutive maxima.
Wavelength Calculation
Wavelength calculation is integral to understanding diffraction patterns because it helps determine the angles at which maxima occur. Using the relationship from the diffraction grating formula:
  • \( \sin(\theta) = \frac{m \lambda}{d} \)
Here, \( m \) is the order of the maxima, \( \lambda \) represents the wavelength of the incident light, and \( d \) is the grating spacing. In the problem, the wavelength \( 589 \, \text{nm} \) is given. Calculating the diffraction angle involves substituting values for \( m \) and solving for \( \theta \). For instance, for the highest order \( m = 5 \), \( \theta \) is approximately \( 62.3^\circ \). Similarly, for \( m = 4 \) and \( m = 3 \), the respective angles are \( 45.0^\circ \) and \( 32.0^\circ \). These calculations reveal how light of a specific wavelength interacts with a grating, manifesting as bright lines at distinct angles.

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Most popular questions from this chapter

The \(D\) line in the spectrum of sodium is a doublet with wavelengths 589.0 and \(589.6 \mathrm{nm}\). Calculate the minimum number of lines needed in a grating that will resolve this doublet in the second-order spectrum.

The wings of tiger beetles (Fig. \(36-41\) ) are colored by interference due to thin cuticle-like layers. In addition, these layers are arranged in patches that are \(60 \mu \mathrm{m}\) across and produce different colors. The color you see is a pointillistic mixture of thinfilm interference colors that varies with perspective. Approximately what viewing distance from a wing puts you at the limit of resolving the different colored patches according to Rayleigh's criterion? Use \(550 \mathrm{nm}\) as the wavelength of light and \(3.00 \mathrm{~mm}\) as the diameter of your pupil.

A plane wave of wavelength \(590 \mathrm{nm}\) is incident on a slit with a width of \(a=0.40 \mathrm{~mm}\). A thin converging lens of focal length \(+70 \mathrm{~cm}\) is placed between the slit and a viewing screen and focuses the light on the screen. (a) How far is the screen from the lens? (b) What is the distance on the screen from the center of the diffraction pattern to the first minimum?

With a particular grating the sodium doublet \((589.00 \mathrm{nm}\) and \(589.59 \mathrm{nm}\) ) is viewed in the third order at \(10^{\circ}\) to the normal and is barely resolved. Find (a) the grating spacing and (b) the total width of the rulings.

Suppose that the central diffraction envelope of a double-slit diffraction pattern contains 11 bright fringes and the first diffraction minima eliminate (are coincident with) bright fringes. How many bright fringes lie between the first and second minima of the diffraction envelope?

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