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Visible light passes through a diffraction grating that has 900 slits/cm, and the interference pattern is observed on a screen that is 2.50 m from the grating. (a) Is the angular position of the first-order spectrum small enough for sin \(\theta \approx \theta\) to be a good approximation? (b) In the first- order spectrum, the maxima for two different wavelengths are separated on the screen by 3.00 mm. What is the difference in these wavelengths?

Short Answer

Expert verified
(a) Yes, the angle is small enough. (b) The wavelength difference is around 20 nm.

Step by step solution

01

Understand the approximation condition

For the approximation \( \sin \theta \approx \theta \) (in radians) to be valid, \( \theta \) should be small, typically less than 10 degrees. First, calculate the angle \( \theta \) for the first-order spectrum to check the approximation condition.
02

Compute diffraction grating spacing

Convert the given grating density of 900 slits/cm to grating spacing \( d \) in meters. \[ d = \frac{1}{900 \, \text{slits/cm}} = \frac{1}{90000 \, \text{slits/m}}\]
03

Determine the angle \( \theta \) using the first-order approximation

The angle \( \theta \) for the first-order maximum is given by the formula \[sin \theta = \frac{m \lambda}{d}\]Since we are interested in the value of \( \theta \) for small approximation validity, note that without a specific wavelength \( \lambda \), we cannot directly calculate \( \theta \) but will instead reason about the typical values in the visible light spectrum (around 500 nm) to do a comparative analysis.
04

Calculate angular separation for two wavelengths

Calculate the angular separation \( \Delta \theta \) between two wavelengths using \[\Delta \theta = \frac{\Delta y}{L}\]where \( \Delta y = 3.00 \, \text{mm} = 0.003 \, \text{m} \) (given separation on screen) and \( L = 2.5 \, \text{m} \). \[\Delta \theta = \frac{0.003}{2.5}\]
05

Compute wavelength difference

Use the relation \[\Delta \theta = \frac{\Delta \lambda}{d}\]Rearrange to find \[\Delta \lambda = d \cdot \Delta \theta\]Substitute the values of \( d \) from Step 2 and \( \Delta \theta \) from Step 4.

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Key Concepts

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

Angular Position
Angular position in the context of a diffraction grating refers to the angle at which the different orders of maximum intensity—the bright fringes—appear on a viewing screen. This is directly related to how light waves interfere with each other after passing through the grating's slits.

To find the angular position, we use the formula \( \sin \theta = \frac{m \lambda}{d} \), where \( m \) is the order number, \( \lambda \) is the wavelength of the light, and \( d \) is the distance between grating slits. If you have visible light around wavelengths of 500 nm, to find the first-order maximum, you would use \( m=1 \).

For small angles, which are often the case in practical scenarios, the approximation \( \sin \theta \approx \theta \) (in radians) is valid. This makes computation simpler, as you can directly measure or calculate \( \theta \) without complex trigonometric functions. Checking smallness can be done by comparing \( \theta \) against 10 degrees (roughly 0.1745 radians). If \( \theta \) is significantly less, the approximation holds well.
Interference Pattern
When light passes through a diffraction grating, it creates an interference pattern—a series of alternating bright and dark spots or lines on a screen. This occurs due to constructive and destructive interference of light waves.

In constructive interference, the path difference between waves is a multiple of the wavelength, causing bright spots, called maxima. For destructive interference, waves cancel each other out, leading to dark spots, called minima.

The spacing between these fringes (bright or dark spots) on the screen is related to both the wavelength of the light and the distance to the screen. Interference patterns are thus directly tied to the wavelength and are a visual representation of the wave properties of light.

These patterns are noticeable and precisely measurable, making diffraction gratings powerful tools in experiments where determining light properties is essential.
Wavelength Difference
The wavelength difference between two light sources in a diffraction grating setup determines how far apart their corresponding maxima will appear on the screen. To determine the separation, one needs to consider both the distance between the grating and the screen, as well as the specific wavelengths involved.

With a known distance to the screen and a measured separation \( \Delta y \) on the screen between maxima, the angular separation \( \Delta \theta \) can be computed using \( \Delta \theta = \frac{\Delta y}{L} \), where \( L \) is the screen distance.

It's straightforward to relate this angular separation back to a difference in wavelength \( \Delta \lambda \) with \( \Delta \theta = \frac{\Delta \lambda}{d} \), fully utilizing the grating spacing \( d \).

The resulting difference in wavelengths is given by \( \Delta \lambda = d \cdot \Delta \theta \), which allows determinations of wavelength differences from readily observable screen measurements—a useful method in spectroscopy.

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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?

Monochromatic electromagnetic radiation with wavelength \(\lambda\) from a distant source passes through a slit. The diffraction pattern is observed on a screen 2.50 m from the slit. If the width of the central maximum is 6.00 mm, what is the slit width \(a\) if the wavelength is (a) 500 nm (visible light); (b) 50.0 \(\mu\)m (infrared radiation); (c) 0.500 nm (x rays)?

A loudspeaker with a diaphragm that vibrates at 960 Hz is traveling at 80.0 m/s directly toward a pair of holes in a very large wall. The speed of sound in the region is 344 m/s. Far from the wall, you observe that the sound coming through the openings first cancels at \(\pm11.4^\circ\) with respect to the direction in which the speaker is moving. (a) How far apart are the two openings? (b) At what angles would the sound first cancel if the source stopped moving?

Diffraction occurs for all types of waves, including sound waves. High- frequency sound from a distant source with wavelength 9.00 cm passes through a slit 12.0 cm wide. A microphone is placed 8.00 m directly in front of the center of the slit, corresponding to point \(O\) in Fig. 36.5a. The microphone is then moved in a direction perpendicular to the line from the center of the slit to point \(O\). At what distances from \(O\) will the intensity detected by the microphone be zero?

The VLBA (Very Long Baseline Array) uses a number of individual radio telescopes to make one unit having an equivalent diameter of about 8000 km. When this radio telescope is focusing radio waves of wavelength 2.0 cm, what would have to be the diameter of the mirror of a visible-light telescope focusing light of wavelength 550 nm so that the visible-light telescope has the same resolution as the radio telescope?

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