Problem 37
Question
The wavelength of the laser beam used in a compact disc player is \(780 \mathrm{~nm}\). Suppose that a diffraction grating produces first-order tracking beams that are \(1.2 \mathrm{~mm}\) apart at a distance of \(3.0 \mathrm{~mm}\) from the grating. Estimate the spacing between the slits of the grating.
Step-by-Step Solution
Verified Answer
The spacing between the slits is approximately \(2.1 \mu m\).
1Step 1: Convert Units
Firstly, convert the given wavelength from nanometers to meters. Knowing that 1 nm = \(10^{-9}\) meters, the converted wavelength is \(780 \times 10^{-9}\) m.
2Step 2: Understand Diffraction Grating Formula
The diffraction grating formula is given by \(d \sin(\theta) = m \lambda\), where \(d\) is the slit spacing, \(\theta\) is the angle of diffraction, \(m\) is the order number, and \(\lambda\) is the wavelength. The order number, \(m\), is 1 since it is first-order diffraction.
3Step 3: Find the Angle of Diffraction
Calculate the angle \(\theta\) using trigonometric identities. With the diffraction beams \(1.2 \mathrm{~mm}\) apart at \(3.0 \mathrm{~mm}\), the tangent of the angle is given by \(\tan(\theta) = \frac{1.2}{3.0}\). Convert to sine using the relation \(\sin(\theta) = \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}\).
4Step 4: Calculate Slit Spacing
Rearrange the diffraction formula to solve for \(d\): \(d = \frac{m \lambda}{\sin(\theta)}\). Substituting the known values \(m = 1\), \(\lambda = 780 \times 10^{-9} \mathrm{~m}\), and \(\sin(\theta)\) from the previous step, compute \(d\).
5Step 5: Perform Calculation
Calculate \(\tan(\theta) = \frac{1.2}{3.0} = 0.4\). Then \(\sin(\theta) = \frac{0.4}{\sqrt{1+0.4^2}} \approx 0.3714\). Finally, compute \(d = \frac{1 \cdot 780 \times 10^{-9}}{0.3714} \approx 2.1 \times 10^{-6} \mathrm{~m}\).
Key Concepts
Laser WavelengthSlit SpacingAngle of DiffractionFirst-Order Diffraction
Laser Wavelength
When we talk about laser wavelength, we are referring to the distance between successive peaks of the laser light wave. A laser used in devices like CD players often has a specific wavelength because it needs to be precise and consistent. In this exercise, the wavelength of the laser is given as 780 nanometers (nm). To work with this in calculations involving diffraction, you usually need to convert it into meters:
- 1 nanometer = 1 x 10-9 meters.
- Hence, 780 nm = 780 x 10-9 meters.
Slit Spacing
Slit spacing refers to the distance between adjacent slits in a diffraction grating. It's an essential part of the diffraction process because it affects how the light waves interfere with each other. This exercise asks us to find the slit spacing of a diffraction grating that creates specific diffraction patterns. We use the diffraction equation, which is:\[d \sin(\theta) = m \lambda\]In this equation:
- \(d\) is the slit spacing.
- \(\theta\) is the angle of diffraction.
- \(m\) is the order number (1st order in this problem).
- \(\lambda\) is the laser wavelength.
Angle of Diffraction
The angle of diffraction is the angle at which light is diffracted from a grating. This angle can be calculated using the positions where light spots appear. Given in this exercise are beams 1.2 mm apart at a distance of 3.0 mm from the grating. We use trigonometry to find the angle:\[\tan(\theta) = \frac{1.2}{3.0}\]Once we have the tangent, we convert it to sine, as used in the diffraction equation:\[\sin(\theta) = \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}}\]This conversion is necessary to substitute into the diffraction formula to find the slit spacing. Understanding this angle is key to unraveling how the pattern spreads out.
First-Order Diffraction
First-order diffraction refers to the primary pattern produced by a diffraction grating. It is the simplest and usually the most pronounced diffraction pattern, characterized by the order number \(m = 1\) in our equation:\[d \sin(\theta) = m \lambda\]This primary order is crucial as it helps establish a baseline for calculations in detecting effective diffraction spread and grating characteristics. The first-order patterns are generally used because they're the most distinct and provide clear data for understanding the interaction of light with the grating.
- Essential for measuring and calculating precise characteristics of light sources.
- It provides a clear example of wave interference resulting in measurable angles and patterns.
Other exercises in this chapter
Problem 35
A spotlight sends red light (wavelength \(=694.3 \mathrm{nm}\) ) to the moon. At the surface of the moon, which is \(3.77 \times 10^{8} \mathrm{~m}\) away, the
View solution Problem 36
Two concentric circles of light emit light whose wavelength is \(555 \mathrm{~nm}\). The larger circle has a radius of \(4.0 \mathrm{~cm}\), and the smaller cir
View solution Problem 38
The light shining on a diffraction grating has a wavelength of \(495 \mathrm{nm}\) (in vacuum). The grating produces a second-order bright fringe whose position
View solution Problem 39
For a wavelength of \(420 \mathrm{~nm}\), a diffraction grating produces a bright fringe at an angle of \(26^{\circ}\). For an unknown wavelength, the same grat
View solution