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Chapter 34: Problem 51

Light from an argon laser strikes a diffraction grating that has 7020 grooves per centimeter. The central and firstorder principal maxima are separated by \(0.332 \mathrm{~m}\) on a wall \(1.00 \mathrm{~m}\) from the grating. Determine the wavelength of the laser light.

Short Answer

Expert verified
Answer: The wavelength of the laser light is approximately \(454 \mathrm{~nm}\).

Step by step solution

01

Find the Grating Separation (d)

The grating separation (d) can be determined by the formula: \(d = \frac{1}{N}\) Where N is the number of grooves per centimeter. In this case, N = 7020 grooves per centimeter. \(d = \frac{1}{7020}\) \(d = 1.42 \times 10^{-4} \mathrm{~cm}\)
02

Convert Grating Separation to Meters

To work with meters in the rest of the problem, convert the grating separation (d) to meters: \(d = 1.42\times10^{-4}\ \text{cm}\times\frac{1\text{ m}}{100\text{ cm}} = 1.42\times10^{-6}\ \text{m}\)
03

Calculate the Angle Between Central and First-Order Maxima (θ)

We are given the distance between the central and first-order maxima (y = 0.332 m) and the distance from the grating to the wall (L = 1.00 m). We can use the right triangle relationship to find the angle θ: \(\tan \theta = \frac{y}{L}\) \(\theta = \arctan{\left(\frac{0.332 \ \text{m}}{1.00 \ \text{m}}\right)}\) \(\theta \approx 18.56^{\circ}\)
04

Calculate the Wavelength Using the Grating Equation

Now, we can use the grating equation to find the wavelength (λ): \(m \lambda = d \sin \theta\) Where m is the order of the principal maximum (m = 1 for the first-order maximum). \(\lambda = \frac{d \sin \theta}{m}\) \(\lambda = \frac{1.42 \times 10^{-6} \ \text{m} \cdot \sin(18.56^{\circ})}{1}\) \(\lambda \approx 4.54 \times 10^{-7} \text{ m}\)
05

Convert Wavelength to Nanometers

To express the wavelength in nanometers, convert the result to nanometers: \(\lambda = 4.54\times10^{-7} \text{ m}\times\frac{10^9\text{ nm}}{1\text{ m}} \approx 454\ \text{nm}\) The wavelength of the laser light is approximately \(454 \mathrm{~nm}\).

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Most popular questions from this chapter

An instructor uses light of wavelength \(633 \mathrm{nm}\) to create a diffraction pattern with a slit of width \(0.135 \mathrm{~mm} .\) How far away from the slit must the instructor place the screen in order for the full width of the central maximum to be \(5.00 \mathrm{~cm} ?\)

Coherent monochromatic light with wavelength \(\lambda=514 \mathrm{nm}\) is incident on two slits that are separated by a distance \(d=0.500 \mathrm{~mm} .\) The intensity of the radiation at a screen \(2.50 \mathrm{~m}\) away from each slit is \(180.0 \mathrm{~W} / \mathrm{cm}^{2} .\) Deter-

Can light pass through a single slit narrower than its wavelength? If not, why not? If so, describe the distribution of the light beyond the slit.

When using a telescope with an objective of diameter \(12.0 \mathrm{~cm},\) how close can two features on the Moon be and still be resolved? Take the wavelength of light to be \(550 \mathrm{nm}\), near the center of the visible spectrum.

A red laser pointer with a wavelength of \(635 \mathrm{nm}\) shines on a diffraction grating with 300 lines \(/ \mathrm{mm}\). A screen is then placed a distance of \(2.0 \mathrm{~m}\) behind the diffraction grating to observe the diffraction pattern. How far away from the central maximum will the next bright spot be on the screen? a) \(39 \mathrm{~cm}\) c) \(94 \mathrm{~cm}\) e) \(9.5 \mathrm{~m}\) b) \(76 \mathrm{~cm}\) d) \(4.2 \mathrm{~m}\)
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