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Light with a wavelength of \(692 \mathrm{~nm}\) shines on a diffraction grating with a slit spacing of \(1.92 \times 10^{-6} \mathrm{~m}\). What is the angle to the second-order principal maximum \((m=2)\) ?

Short Answer

Expert verified
46.13°

Step by step solution

01

Understand the Formula

The formula to find the angle of the principal maximum in a diffraction grating is given by \( d \sin \theta = m \lambda \), where \( d \) is the spacing between the slits, \( \theta \) is the angle, \( m \) is the order of the maximum, and \( \lambda \) is the wavelength of the light.
02

Plug in the Given Values

In this problem, we have \( \lambda = 692 \) nm which is \( 692 \times 10^{-9} \) meters, \( d = 1.92 \times 10^{-6} \) meters, and \( m = 2 \). We plug these values into the formula: \( 1.92 \times 10^{-6} \times \sin \theta = 2 \times 692 \times 10^{-9} \).
03

Solve for \( \sin \theta \)

Rearrange the equation to solve for \( \sin \theta \):\[ \sin \theta = \frac{2 \times 692 \times 10^{-9}}{1.92 \times 10^{-6}} \].Calculate this value to find \( \sin \theta \).
04

Calculate \( \sin \theta \)

Perform the division:\( \sin \theta = \frac{1384 \times 10^{-9}}{1.92 \times 10^{-6}} = 0.7208 \) (approximately).
05

Find the Angle \( \theta \)

Use the inverse sine function to find \( \theta \):\( \theta = \sin^{-1}(0.7208) \).Calculate this to find \( \theta \).
06

Final Calculation

Using a calculator, \( \theta = \sin^{-1}(0.7208) \approx 46.13° \).
07

Conclusion

The angle to the second-order principal maximum, \( m = 2 \), is approximately \( 46.13° \).

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Key Concepts

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

Wavelength
Wavelength is a fundamental concept in the study of waves, particularly electromagnetic waves such as light. It defines the distance between successive peaks or troughs of a wave. In the context of a diffraction grating, the wavelength of light determines how the light will interact with the slits of the grating. Light has different colors because of different wavelengths, and these variations affect how light diffracts around obstacles and openings.

In our example, the light with a wavelength of 692 nanometers (nm) is being used. Firstly, it's important to convert this wavelength from nanometers to meters for proper calculations, as the diffraction grating's spacing is typically given in meters. Thus, 692 nm equals 692 x 10^-9 meters. This conversion is necessary for further calculations when applying the diffraction formula.
Principal Maximum
The principal maximum refers to the prominent bright spots you see on a screen when light passes through a diffraction grating. These spots are the result of constructive interference, where multiple light waves meet in phase and their amplitudes add up.

Constructive interference occurs at specific angles where the path difference between adjacent waves from different slits is exactly an integer multiple of the wavelength. The order of the principal maximum (denoted by 'm') specifies this integer multiple. For instance, the first-order maximum (m=1) occurs where the path difference is one wavelength, the second-order (m=2) with two wavelengths, and so on.

This principle is crucial in determining the angle at which these maxima occur, using the formula related to diffraction gratings.
Angle Calculation
Calculating the angle to a principal maximum involves using the formula for diffraction gratings: \[ d \sin \theta = m \lambda \] Here, \( d \) is the distance between slits, \( \theta \) is the angle of diffraction, \( m \) is the order of the maximum, and \( \lambda \) is the wavelength. This mathematical relation explains how the light waves interfere to form maxima at specific angles.

For our exercise, to find the angle of the second-order principal maximum, we rearrange the formula to solve for \( \sin \theta \). Plugging in the given values: \( d = 1.92 \times 10^{-6} \) meters, \( \lambda = 692 \times 10^{-9} \) meters, and \( m = 2 \), the formula becomes: \[ \sin \theta = \frac{2 \times 692 \times 10^{-9}}{1.92 \times 10^{-6}} \approx 0.7208 \] The inverse sine function then helps us find the angle \( \theta \), resulting in approximately \( 46.13° \).

This calculated angle corresponds to the second-order brightness seen on a diffraction pattern.
Order of Maximum
The order of maximum, indicated by \( m \), is an essential factor in diffraction grating calculations. It signifies which principal maximum you are analyzing. For instance, \( m=1 \) corresponds to the first bright fringe and \( m=2 \) represents the second, and so forth.

With each increase in the order of maximum, the angle \( \theta \) typically becomes larger as the path difference increases, hence the need for higher order calculations like \( m=2 \) in our case. This is why calculating the exact angle to the maximum is important; it helps in identifying which bright fringe appears at which angle.

Moreover, higher orders might not always be visible due to limitations in the diffraction setup such as angle constraints or the physical properties of the grating, but they can be predicted and calculated using the formula. Understanding the order of maximum is fundamental to predicting and analyzing patterns created by light passing through diffraction gratings.

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Most popular questions from this chapter

Two plates of glass are separated by an air-filled space with a thickness of \(5.1 \times 10^{-5} \mathrm{~m}\). What wavelength of light experiences constructive interference for \(m=250\) ?

A soap film ( \(n=1.33\) ) is \(825 \mathrm{~nm}\) thick. What wavelengths in the range from \(400 \mathrm{~nm}\) to \(700 \mathrm{~nm}\) (visible light) will experience destructive interference if the film is surrounded by air on both sides?

A thin film of oil ( \(n=1.35\) ) floats on a puddle of water ( \(n=1.33\) ). If light with a wavelength of \(590 \mathrm{~nm}\) experiences destructive interference when it reflects from the film, what is the minimum thickness of the oil?

A physics instructor wants to produce a two-slit interference pattern large enough for her class to see, so she projects the pattern onto a screen. She would like the distance on the screen between the central bright fringe and the bright fringe just above it to be \(2.50 \mathrm{~cm}\). If the slits have a separation of \(d=0.0220 \mathrm{~mm}\), what is the required distance from the slits to the screen when \(632.8-\mathrm{nm}\) light from a helium-neon laser is used?

A two-slit experiment with red light produces a set of bright fringes. (a) Will the spacing between the fringes increase, decrease, or stay the same if the color of the light is changed to blue? (b) Choose the best explanation from among the following: A. The spacing will increase because blue light has a greater frequency than red light. B. The spacing will decrease because blue light has a smaller wavelength than red light. C. The spacing will stay the same because only the wave nature of light is important, not the color of the light.

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