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Chapter 27: Problem 49

A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and 660 \(\mathrm{nm}\) . What is (are) the wavelength(s) of the incident light that could have produced this maximum?

Short Answer

Expert verified
The wavelength is 640 nm.

Step by step solution

01

Understand the Grating Equation

The diffraction grating equation is \( d \sin \theta = m\lambda \), where \( d \) is the distance between lines on the grating, \( \theta \) is the angle of diffraction, \( m \) is the order of the spectrum, and \( \lambda \) is the wavelength. We need to calculate \( d \) from the given number of lines per centimeter.
02

Calculate Line Spacing

The spacing \( d \) is the inverse of the number of lines per centimeter. So, \( d = \frac{1 \text{ cm}}{2604} \). First, convert it to meters: \( d = \frac{1 \text{ m}}{260400} = 3.84 \times 10^{-6} \text{ m} \).
03

Applying the Grating Equation

For principal maximum at \( \theta=30^{\circ} \), plug values into the equation: \( 3.84 \times 10^{-6} \sin 30^{\circ} = m\lambda \). Simplify using \( \sin 30^{\circ} = 0.5 \): \( 1.92 \times 10^{-6} = m\lambda \).
04

Solve for Wavelength (\( \lambda \))

Rearrange to find \( \lambda = \frac{1.92 \times 10^{-6}}{m} \). Given \( \lambda \) must be between 410 nm (\( 410 \times 10^{-9} \, \text{m} \)) and 660 nm (\( 660 \times 10^{-9} \, \text{m} \)), check which integer values of \( m \) make \( \lambda \) fall within this interval.
05

Calculate Valid m and \( \lambda \) Values

Calculate \( m = 1, 2, 3, ... \) and check: - For \( m=1 \), \( \lambda = \frac{1.92 \times 10^{-6}}{1} = 1920 \times 10^{-9} \, \text{m} \) (too high)- For \( m=2 \), \( \lambda = \frac{1.92 \times 10^{-6}}{2} = 960 \times 10^{-9} \, \text{m} \) (too high)- For \( m=3 \), \( \lambda = \frac{1.92 \times 10^{-6}}{3} = 640 \times 10^{-9} \, \text{m} \) (valid)Thus, for \( m=3 \), \( \lambda = 640 \text{ nm} \).
06

Conclude the Wavelengths

The wavelength of the incident light that could have produced the maximum is 640 nm.

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Key Concepts

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

Grating Equation
The grating equation is a fundamental principle that is used to understand how light interacts with a diffraction grating. The equation is given by \( d \sin \theta = m\lambda \), where:
  • \( d \) represents the distance between adjacent lines on the grating.
  • \( \theta \) is the angle at which light is diffracted.
  • \( m \) denotes the order of the spectrum, an integer value that indicates the number of wavelengths by which paths differ from adjacent slits.
  • \( \lambda \) is the wavelength of the light.
This equation helps us predict the angles at which we will observe constructive interference (bright spots) of light, creating a diffraction pattern. These patterns depend on the wavelength of light being used and the spacing of the grating lines. Understanding this equation is crucial for solving problems related to diffraction gratings.
Line Spacing Calculation
Line spacing in a diffraction grating refers to the distance between adjacent lines, and it's a key variable in the grating equation. In our exercise, the diffraction grating has 2604 lines per centimeter. To calculate the line spacing \( d \), we take the reciprocal of the number of lines per unit length. Convert lines per centimeter to lines per meter for easier calculation:
  • \( \text{lines per meter} = 2604 \times 100 = 260400 \text{ lines} \)
Now, calculate \( d \): \[ d = \frac{1 \text{ meter}}{260400 \text{ lines}} = 3.84 \times 10^{-6} \text{ meters} \] This small value of \( d \) allows us to effectively use the grating equation to study how light of different wavelengths interacts with the grating.
Wavelength Determination
Determining the wavelength \( \lambda \) that could produce a specific diffracted angle involves using the grating equation once we know the line spacing \( d \) and the diffracted angle \( \theta \). For this exercise, \( \theta = 30^{\circ} \). The rearranged grating equation is: \[ \lambda = \frac{d \sin \theta}{m} \] Plugging in the known values:
  • \( d = 3.84 \times 10^{-6} \text{ meters} \)
  • \( \sin 30^{\circ} = 0.5 \)
This becomes: \[ \lambda = \frac{3.84 \times 10^{-6} \times 0.5}{m} = \frac{1.92 \times 10^{-6}}{m} \text{ meters} \] To determine which wavelengths are possible, understand that \( \lambda \) must fall between 410 nm and 660 nm. We evaluate integer values for \( m \) to find valid wavelengths.
Order of Spectrum
The term "order of spectrum" \( m \) in the context of a diffraction grating refers to the various diffracted beams of light that are observed. Each order relates to different multiples of wavelengths. Only certain orders will result in wavelengths within the given range, as highlighted in our exercise. To find which orders are feasible:
  • Begin with \( m = 1 \), resulting in \( \lambda = \frac{1.92 \times 10^{-6}}{1} = 1920 \text{ nm} \), which is outside the range.
  • \( m = 2 \) results in \( \lambda = \frac{1.92 \times 10^{-6}}{2} = 960 \text{ nm} \), also outside the range.
  • Finding \( m = 3 \) gives \( \lambda = \frac{1.92 \times 10^{-6}}{3} = 640 \text{ nm} \), which is valid and within range.
Through this method, we identify that the third order, \( m = 3 \), is the one where the wavelength of the incident light producing the maximum is 640 nm.

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Most popular questions from this chapter

A hunter who is a bit of a braggart claims that from a distance of 1.6 km he can selectively shoot either of two squirrels who are sitting ten centimeters apart on the same branch of a tree. What鈥檚 more, he claims that he can do this without the aid of a telescopic sight on his rifle. (a) Determine the diameter of the pupils of his eyes that would be required for him to be able to resolve the squirrels as separate objects. In this calculation use a wavelength of 498 nm (in vacuum) for the light. (b) State whether his claim is reasonable, and provide a reason for your answer. In evaluating his claim, consider that the human eye automatically adjusts the diameter of its pupil over a typical range of 2 to 8 mm, the larger values coming into play as the lighting becomes darker. Note also that under dark conditions, the eye is most sensitive to a wavelength of 498 nm.

A circular drop of oil lies on a smooth, horizontal surface. The drop is thickest in the center and tapers to zero thickness at the edge. When illuminated from above by blue light \((\lambda=455 \mathrm{nm}), 56\) concentric bright rings are visible, including a bright fringe at the edge of the drop. In addition, there is a bright spot in the center of the drop. When the drop is illuminated from above by red light \((\lambda=637 \mathrm{nm})\) a bright spot again appears at the center, along with a different number of bright rings. Ignoring the bright spot, how many bright rings appear in red light? Assume that the index of refraction of the oil is the same for both wavelengths.

A nonreflective coating of magnesium fluoride \((n=1.38)\) covers the glass \((n=1.52)\) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum \(=565 \mathrm{nm}\) ), determine the minimum nonzero thickness that the coating can have.

When monochromatic light shines perpendicularly on a soap film \((n=1.33)\) with air on each side, the second smallest nonzero film thickness for which destructive interference of reflected light is observed is 296 \(\mathrm{nm}\) . What is the vacuum wavelength of the light in \(\mathrm{nm}\) ?

A large group of football fans comes to the game with colored cards that spell out the name of their team when held up simultaneously. Most of the cards are colored blue \(\left(\lambda_{\text { vacum }}=480 \mathrm{nm}\right)\) . When displayed, the average distance between neighboring cards is 5.0 \(\mathrm{cm} .\) If the cards are to blur together into solid blocks of color when viewed by a spectator at the other end of the stadium \((160 \mathrm{m}\) away), what must be the maximum diameter (in \(\mathrm{mm}\) ) of the spectator's pupils?
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