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At what angle will the second-order maximum be seen from a diffraction grating of spacing \(1.25 \mu \mathrm{m}\) when illuminated by light of wavelength \(550 \mathrm{nm} ?\)

Short Answer

Expert verified
The second-order maximum is seen at an angle of approximately \(61.59^\circ\).

Step by step solution

01

Convert Units

First, convert all the given measurements into consistent units for calculation purposes. The wavelength is given as \(550 \mathrm{nm}\), which can be converted to meters as follows: \(550 \mathrm{nm} = 550 \times 10^{-9} \mathrm{m}\). The grating spacing is given as \(1.25 \mu \mathrm{m}\), which can be converted to meters as follows: \(1.25 \mu \mathrm{m} = 1.25 \times 10^{-6} \mathrm{m}\).
02

Identify the Formula

The formula to find the angle at which the diffraction maxima occur is given by the diffraction grating equation: \[ d \sin \theta = n \lambda \] where \(d\) is the grating spacing, \(\theta\) is the angle, \(n\) is the order of the maximum, and \(\lambda\) is the wavelength of the light.
03

Substitute Known Values

Substitute the known values into the equation \(d \sin \theta = n \lambda\). Here, \(d = 1.25 \times 10^{-6} \mathrm{m}\), \(n = 2\), and \(\lambda = 550 \times 10^{-9} \mathrm{m}\). Thus, the equation becomes: \[ 1.25 \times 10^{-6} \sin \theta = 2 \times 550 \times 10^{-9} \]
04

Solve for \(\sin \theta\)

Rearrange the equation to solve for \(\sin \theta\): \[ \sin \theta = \frac{2 \times 550 \times 10^{-9}}{1.25 \times 10^{-6}} \] Calculate the value: \[ \sin \theta = \frac{1100 \times 10^{-9}}{1.25 \times 10^{-6}} = 0.88 \]
05

Calculate \(\theta\)

Find \(\theta\) by taking the inverse sine of \(0.88\). Use a calculator to find: \[ \theta = \sin^{-1}(0.88) \] Therefore, \(\theta \approx 61.59^\circ\).

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Key Concepts

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

Wavelength Conversion
When dealing with problems involving diffraction, it's crucial to make sure all units are consistent. Wavelength is often given in nanometers (nm), especially when referring to visible light. Visible light typically ranges from about 380 nm to 750 nm. However, converting these measurements to meters makes calculations compatible and easier.

For instance, if you have a wavelength of 550 nm, you convert it to meters since most equations in physics, including the diffraction grating equation, are based on the metric system. The conversion factor is straightforward:
  • 1 nanometer (nm) = 1 x 10鈦烩伖 meters (m).
Consequently, a wavelength of 550 nm is equal to 550 x 10鈦烩伖 meters. Similarly, grating spacing often provided in micrometers (碌m) should also be converted:
  • 1 micrometer (碌m) = 1 x 10鈦烩伓 meters (m).
Ensuring both wavelength and spacing are in meters allows the use of these values seamlessly into the diffraction grating formula.
Diffraction Angle
The diffraction angle is an important concept in understanding how light interacts with a diffraction grating. When light passes through a diffraction grating, it is split into different angles. These angles determine how the light spreads out.

The angle is dependent on several factors: the order of the maximum, the wavelength of the light, and the spacing of the grating. The key equation that describes this is: \[ d \sin \theta = n \lambda \]
  • \(d\) is the spacing of the diffraction grating,
  • \(\theta\) is the angle of diffraction,
  • \(n\) is the order of the maximum,
  • \(\lambda\) is the wavelength.
The angle \(\theta\) is measured from the normal, which is a line perpendicular to the surface of the grating.

This angle tells us where the light beam will form its brightest spots or maxima. In the specific problem, \(\theta\) was calculated to be about 61.59 degrees for the second-order maximum.
Diffraction Maximum
A diffraction maximum is one of the bright spots created when light passes through a diffraction grating. The maxima occur at specific angles where the light waves from different slits add constructively.

When we talk about diffraction maxima, we're referring to points where the light intensity is higher due to constructive interference. Constructive interference happens when the path difference between the waves is a whole number of wavelengths, dictated by the order of the maximum \(n\).
  • The zero-order maximum \(n = 0\) is essentially the directly transmitted light.
  • First-order maxima \(n = 1\) are the first bright spots on either side of the zero-order maximum.
  • Second-order maxima \(n = 2\), third-order \(n = 3\), and so on occur at greater angles as \(n\) increases.
Using the diffraction equation \[ d \sin \theta = n \lambda \], you can calculate at which angle any-order maximum will occur. For example, when \(n = 2\), the second-order maximum is calculated and found to be at an angle of approximately 61.59 degrees. Understanding this helps in determining how different wavelengths are separated by a grating.

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Most popular questions from this chapter

In a double-slit experiment that uses monochromatic light, the angular separation between the central maximum and the second-order maximum is \(0.160^{\circ} .\) What is the wavelength of the light if the distance between the slits is \(0.350 \mathrm{~mm} ?\)

White light whose components have wavelengths from \(400 \mathrm{nm}\) to \(700 \mathrm{nm}\) illuminates a diffraction grating with 4000 lines \(/ \mathrm{cm} .\) Do the first- and second-order spectra overlap? Justify your answer.

(a) In a double-slit experiment, if the distance from the double slits to the screen is increased, the separation between the adiacent maxima will (1) increase, (2) decrease, (3) remain the same. Explain. (b) Yellow-green light \((\lambda=550 \mathrm{nm})\) illuminates a double-slit separated by \(1.75 \times 10^{-4} \mathrm{~m} .\) If the screen is located \(2.00 \mathrm{~m}\) from the slits, determine the separation between the adjacent maxima. (c) What if the screen is located \(3.00 \mathrm{~m}\) from the slits?

Light of two different wavelengths is used in a double-slit experiment. The location of the third-order maximum for the first light, yellow-orange light \((\lambda=600 \mathrm{nm})\) coincides with the location of the fourth-order maximum for the other color's light. What is the wavelength of the other light?

Two parallel slits are illuminated with monochromatic light, and an interference pattern is observed on a screen. (a) If the distance between the slits were decreased, would the distance between the maxima (1) increase, (2) remain the same, or (3) decrease? Explain. (b) If the slit separation is \(1.0 \mathrm{~mm}\), the wavelength is \(640 \mathrm{nm}\), and the distance from the slits to the screen is \(3.00 \mathrm{~m}\), what is the separation between adjacent interference maxima? (c) What if the slit separation is \(0.80 \mathrm{~mm} ?\)

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