/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 44 Perhaps to confuse a predator, s... [FREE SOLUTION] | 91Ó°ÊÓ

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Perhaps to confuse a predator, some tropical gyrinid beetles (whirligig beetles) are colored by optical interference that is due to scales whose alignment forms a diffraction grating (which scatters light instead of transmitting it). When the incident light rays are perpendicular to the grating, the angle between the firstorder maxima (on opposite sides of the zeroth-order maximum) is about \(26^{\circ}\) in light with a wavelength of \(550 \mathrm{nm}\). What is the grating spacing of the beetle?

Short Answer

Expert verified
The grating spacing is approximately 2455 nm.

Step by step solution

01

Identify Given Data

First, take note of the values provided in the problem. The angle between the first-order maxima (\( \theta \)) is \(13^{\circ}\) on each side since the total angle between them is \(26^{\circ}\), and the wavelength (\( \lambda \)) of light is given as \(550 \, \text{nm} = 550 \times 10^{-9} \text{ m}\).
02

Understand the Formula

Use the diffraction grating equation which is \( d \sin \theta = m \lambda \), where \(d\) is the grating spacing, \(\theta\) is the angle, \(m\) is the order of the maximum (in this case, 1), and \(\lambda\) is the wavelength.
03

Solve for Grating Spacing

Rearrange the formula to solve for \(d\): \[ d = \frac{m \lambda}{\sin \theta} \].Substitute the known values: \(m = 1\), \(\lambda = 550 \times 10^{-9} \text{ m}\), and \(\theta = 13^{\circ}\).
04

Calculate \(\sin \theta\)

Use a calculator to find \(\sin 13^{\circ}\): \(\sin 13^{\circ} \approx 0.224\).
05

Compute Grating Spacing

Insert the values into the rearranged diffraction equation: \[ d = \frac{1 \times 550 \times 10^{-9}}{0.224} \approx 2455 \times 10^{-9} \text{ m} \].
06

Conclude the Calculation

Simplify the expression: \(d \approx 2455 \text{ nm} \). Therefore, the grating spacing of the beetle is approximately 2455 nm.

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Key Concepts

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

Optical Interference
Optical interference occurs when two or more light waves overlap and combine to form a new wave pattern. This fascinating phenomenon is based on the principle of superposition, meaning that the resultant wave is a simple sum of the individual waves.
In nature, optical interference is responsible for the vibrant colors seen on the wings of beetles, butterflies, and even some sea shells. This happens because these structures have microstructures that act like a diffraction grating.
When light hits these microstructures, the pattern of reflected light waves results in interference. In some cases, certain wavelengths of light (colors) are reinforced due to constructive interference, while others are canceled out due to destructive interference.
  • Constructive interference happens when waves align to amplify the light.
  • Destructive interference occurs when waves align in such a way that they diminish or even completely cancel the light.
These interference patterns result in what we recognize as the iridescence seen on the beetles in the original problem. It is a natural example of how diffraction grating can create beautiful optical effects.
Grating Spacing Calculation
To solve problems involving diffraction gratings, understanding how to calculate grating spacing is crucial. A diffraction grating consists of many evenly spaced lines or grooves that diffract light into several beams.
This grating works because each groove reflects the light wave at a specific angle, based on the angle at which the light hits the grating and the light's wavelength. The key to finding the grating spacing lies in the diffraction equation: \[ d \sin \theta = m \lambda \]Where:
  • \(d\) is the grating spacing to be calculated.
  • \(\theta\) is the angle of the diffracted light.
  • \(m\) is the order of the maxima (first order, in this case).
  • \(\lambda\) is the light's wavelength.
By rearranging this formula to solve for \(d\), you can calculate the spacing:\[ d = \frac{m \lambda}{\sin \theta} \]In the original problem, all required values such as wavelength and angle were provided, allowing for direct calculation of the grating spacing of the beetle’s scales.
Wavelength of Light
The wavelength of light is a critical factor in understanding optical phenomena like diffraction and interference. Wavelength is the distance between consecutive peaks of a light wave, expressed in units of length such as nanometers (nm).
A broad spectrum of wavelengths makes up visible light, each one corresponding to a different color that we perceive.
  • Red light has longer wavelengths, typically around 620-750 nm.
  • Violet light has shorter wavelengths, around 380-450 nm.
In the given problem, the wavelength of interest was 550 nm, typical of green light, which falls in the middle of the visible spectrum. This specific wavelength interacted with the beetle's diffraction grating to produce observable first-order maxima at a specific angle, illustrating the intricate dance between light wavelength and diffraction processes.
Understanding wavelength allows scientists and engineers to predict and analyze light behavior when interacting with different materials and structures, enabling innovations in fields like optical engineering and design.

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Most popular questions from this chapter

The pupil of a person's eye has a diameter of \(5.00 \mathrm{~mm}\). According to Rayleigh's criterion, what distance apart must two small objects be if their images are just barely resolved when they are \(250 \mathrm{~mm}\) from the eye? Assume they are illuminated with light of wavelength \(500 \mathrm{nm}\).

A slit \(1.00 \mathrm{~mm}\) wide is illuminated by light of wavelength \(589 \mathrm{nm}\). We see a diffraction pattern on a screen \(3.00 \mathrm{~m}\) away. What is the distance between the first two diffraction minima on the same side of the central diffraction maximum?

Sound waves with frequency \(3000 \mathrm{~Hz}\) and speed \(343 \mathrm{~m} / \mathrm{s}\) diffract through the rectangular opening of a speaker cabinet and into a large auditorium of length \(d=100 \mathrm{~m}\). The opening, which has a horizontal width of \(30.0 \mathrm{~cm},\) faces a wall \(100 \mathrm{~m}\) away (Fig. \(36-36\) ). Along that wall, how far from the central axis will a listener be at the first diffraction minimum and thus have difficulty hearing the sound? (Neglect reflections.)

In the single-slit diffraction experiment of Fig. \(36-4,\) let the wavelength of the light be \(500 \mathrm{nm}\), the slit width be \(6.00 \mu \mathrm{m}\), and the viewing screen be at distance \(D=3.00 \mathrm{~m}\). Let a \(y\) axis extend upward along the viewing screen, with its origin at the center of the diffraction pattern. Also let \(I_{P}\) represent the intensity of the diffracted light at point \(P\) at \(y=15.0 \mathrm{~cm} .\) (a) What is the ratio of \(I_{P}\) to the intensity \(I_{m}\) at the center of the pattern? (b) Determine where point \(P\) is in the diffraction pattern by giving the maximum and minimum between which it lies, or the two minima between which it lies.

Monochromatic light (wavelength \(=450 \mathrm{nm}\) ) is incident perpendicularly on a single slit (width \(=0.40 \mathrm{~mm}\) ). A screen is placed parallel to the slit plane, and on it the distance between the two minima on either side of the central maximum is \(1.8 \mathrm{~mm}\). (a) What is the distance from the slit to the screen? (Hint: The angle to either minimum is small enough that \(\sin \theta \approx \tan \theta .)\) (b) What is the distance on the screen between the first minimum and the third minimum on the same side of the central maximum?

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