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Chapter 26: Problem 37

\(\bullet\) A diffraction grating has 5580 lines/cm. When a beam of monochromatic light goes through it, the second pair of bright spots occurs at 26.3 \(\mathrm{cm}\) from the central spot on a screen 42.5 \(\mathrm{cm}\) past the grating. (a) What is the wavelength of this light? (b) How far from the central spot does the next pair of bright spots occur on the screen?

Short Answer

Expert verified
(a) The wavelength is 473 nm. (b) The next pair of bright spots is 69 cm from the central spot.

Step by step solution

01

Calculate the Grating Line Spacing

The diffraction grating has 5580 lines per cm. To find the spacing between the lines, called the grating constant \(d\), we use the formula:\[ d = \frac{1}{N} \]where \(N\) is the number of lines per cm. Plugging in the values:\[ d = \frac{1}{5580} \text{ cm} = \frac{1}{558000} \text{ m} \]
02

Determine the Diffraction Angle

In a diffraction setup, the angle \(\theta\) specifies the direction of the bright spots created by a diffraction grating. Using geometry, \(\tan(\theta)\) is given by:\[ \tan(\theta) = \frac{y}{L} \]where \(y\) is the distance from the central maximum to the bright spot (26.3 cm or 0.263 m), and \(L\) is the distance from the grating to the screen (42.5 cm or 0.425 m). Thus:\[ \tan(\theta) = \frac{0.263}{0.425} \approx 0.6188 \]Solving for \(\theta\), we find:\[ \theta = \tan^{-1}(0.6188) \approx 31.8^\circ \]
03

Calculate the Wavelength of the Light

Using the diffraction grating equation for the second-order maximum (\(m = 2\)):\[ d \sin(\theta) = m\lambda \]Substituting the known values (\(d = \frac{1}{558000} \text{ m}\), \(\theta = 31.8^\circ\), and \(m = 2\)), we solve for \(\lambda\):\[ \lambda = \frac{d \sin(31.8^\circ)}{2} \]Subscript \(\sin(31.8^\circ)\) into the equation:\[ \lambda = \frac{\left(\frac{1}{558000}\right) \cdot 0.5274}{2} \approx 4.73 \times 10^{-7} \text{ m} = 473 \text{ nm} \]
04

Find the Position of the Next Pair of Bright Spots

The next pair of bright spots corresponds to the third-order maximum (\(m = 3\)). We need to find \(\theta_3\) where:\[ d \sin(\theta_3) = 3\lambda \]Rearrange to solve for \(\theta_3\):\[ \sin(\theta_3) = \frac{3\lambda}{d} = \frac{3 \cdot 4.73 \times 10^{-7}}{\frac{1}{558000}} \approx 0.791 \]Substitute back to find \(\theta_3\):\[ \theta_3 = \sin^{-1}(0.791) \approx 52.2^\circ \]Now, we calculate \(y_3\), which is the distance from the central maximum for the third-order spots using:\[ \tan(\theta_3) = \frac{y_3}{L} \Rightarrow y_3 = L \tan(\theta_3) \]Substitute and solve for \(y_3\):\[ y_3 = 0.425 \times \tan(52.2^\circ) \approx 0.690 \text{ m} = 69 \text{ cm} \]

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Key Concepts

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

Grating Equation
In order to fully understand how a diffraction grating works, we need to look at the grating equation. This equation helps us determine the angles at which light of different wavelengths will be diffracted. The grating equation is given by:\[ d \sin(\theta) = m\lambda \]Where:
  • d is the grating constant or line spacing, which is the distance between adjacent lines on the grating.
  • \theta is the angle at which light is diffracted.
  • m is the order of the diffraction maximum (1 for the first-order, 2 for the second-order, etc.).
  • \lambda is the wavelength of the incident light.
To solve for the wavelength or another unknown, you'll simply rearrange this equation. The grating equation is essential for predicting where bright spots will appear on a screen.
Wavelength Calculation
Calculating the wavelength of the light is an important step in understanding diffraction patterns produced by a grating. Once we've identified the angle \(\theta\) using the geometry setup between the grating, the slit, and the screen, we can solve for the wavelength using:\[ \lambda = \frac{d \sin(\theta)}{m} \]For example, using the second-order bright spot (where \(m = 2\)), and knowing the line spacing \(d\) and angle \(\theta\) from measurements, we apply the grating equation to find \(\lambda\). After calculating, if you reach a result close to \(473\) nm, you've correctly found the wavelength as occurs in this exercise. This value is typical for visible light, improving understanding of how light interacts with a grating.
Diffraction Line Spacing
Line spacing in a diffraction grating, indicated by \(d\), is the inverse of the number of lines per length (usually given per cm). For instance, if a grating has 5580 lines per cm, the line spacing is calculated as:\[ d = \frac{1}{5580} \text{ cm} = \frac{1}{558000} \text{ m} \]Line spacing is crucial because it determines how the light will diffract when it hits the grating. This means that smaller spacing (more lines per cm) gives rise to a greater spread of the diffraction pattern. As the interval between lines decreases, the angle at which light is spread increases, making precise calculations imperative for experimental success.
Order of Bright Spots
The concept of "order" in diffraction is integral in determining the positions of bright spots or maxima. In practical scenarios, each order corresponds to a set of bright spots spreading outwards from the central maximum. Order is represented as \(m\) in calculations.
  • The 0th order (\(m=0\)) is the central bright spot, directly in front of the grating.
  • The 1st order (\(m=1\)) spot appears at an angle derived from the equation.
  • Higher orders, such as the 2nd (\(m=2\)) and 3rd (\(m=3\)), form further away from the center.
Understanding and identifying these orders aid calculations by allowing for accurate angle differentiation for different wavelengths of light, revealing colored patterns based on interference. In the given exercise, the challenge was to determine the "spots" for the second and third orders, translating calculated angles back to positions on the screen.

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Most popular questions from this chapter

\(\bullet\) A person with a radio-wave receiver starts out equidistant from two FM radio transmitters \(A\) and \(B\) that are 11.0 \(\mathrm{m}\) apart, each one emitting in-phase radio waves at 92.0 \(\mathrm{MH} z\) . She then walks so that she always remains 50.0 \(\mathrm{m}\) from transmitter \(B\) . See Fig- ure \(26.48 .\) ) For what values of \(x\) will she find the radio signal to be (a) maximally enhanced, (b) can- celled? Limit your solution to the cases where \(x \geq 50.0 \mathrm{m.}\)

\( (f鈥� # \)\bullet$$\bullet\( Two small loudspeakers that are 5.50 \)\mathrm{m}\( apart are emitting sound in phase. From both of them, you hear a singer singing \)\mathrm{C} \\#\( (frequency 277 \)\mathrm{Hz} )\( , while the speed of sound in the room is 340 \)\mathrm{m} / \mathrm{s}\( . Assuming that you are rather far from these speak- ers, if you start out at point \)P\( equidistant from both of them and walk around the room in front of them, at what angles (measured relative to the fine from \)P$ to the midpoint between the speakers) will you hear the sound (a) maximally enhanced, (b) cancelled? Neglect any reflections from the walls.

\(\bullet$$\bullet\) The radius of curvature of the convex surface of a planoconvex lens is 95.2 \(\mathrm{cm} .\) The lens is placed convex side down on a perfectly flat glass plate that is illuminated from above with red light having a wavelength of 580 nm. Find the diameter of the second bright ring in the interference pattern.

\(\bullet\) Radio interference. Two radio antennas \(A\) and \(B\) radiate in phase. Antenna \(B\) is 120 \(\mathrm{m}\) to the right of antenna \(A .\) Consider point \(Q\) along the extension of the line connecting the anten- nas, a horizontal distance of 40 \(\mathrm{m}\) to the right of antenna \(B\) . The frequency, and hence the wavelength, of the emitted waves can be varied. (a) What is the longest wavelength for which there will be destructive interference at point \(Q ?\) (b) What is the longest wavelength for which there will be constructive interference at point \(Q ?\)

\(\bullet$$\bullet\) Sensitive eyes. After an eye examination, you put some eyedrops on your sensitive eyes. The cornea (the front part of the eye) has an index of refraction of 1.38, while the eyedrops have a refractive index of \(1.45 .\) After you put in the drops, your friends notice that your eyes look red, because red light of wavelength 600 \(\mathrm{nm}\) has been reinforced in the reflected light. (a) What is the minimum thickness of the film of eyedrops on your cornea? (b) Will any other wavelengths of visible light be reinforced in the reflected light? Will any be cancelled? (c) Suppose you had contact lenses, so that the eyedrops went on them instead of on your corneas. If the refractive index of the lens material is 1.50 and the layer of eyedrops has the same thickness as in part (a), what wavelengths of visible light will be reinforced? What wavelengths will be cancelled?
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