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Chapter 34: Problem 34

The Michelson interferometer is used in a class of commercially available optical instruments called wavelength meters. In a wavelength meter, the interferometer is illuminated simultaneously with the parallel beam of a reference laser of known wavelength and that of an unknown laser. The movable mirror of the interferometer is then displaced by a distance \(\Delta d,\) and the number of fringes produced by each laser and passing by a reference point (a photo detector) is counted. In a given wavelength meter, a red He-Ne laser \(\left(\lambda_{\mathrm{Red}}=632.8 \mathrm{nm}\right)\) is used as a reference laser. When the movable mirror of the interferometer is displaced by a distance \(\Delta d\), a number \(\Delta N_{\text {Red }}=6.000 \cdot 10^{4}\) red fringes and \(\Delta N_{\text {unknown }}=7.780 \cdot 10^{4}\) fringes pass by the reference photodiode. a) Calculate the wavelength of the unknown laser. b) Calculate the displacement, \(\Delta d\), of the movable mirror.

Short Answer

Expert verified
\(\Delta d = \frac{1000 * 632.8 * 10^{-9}}{2}\) Calculating, we get: \(\Delta d = 316.4 * 10^{-6}\) meters Now, we can use this calculated displacement \(\Delta d\) and the number of fringes produced by the unknown laser to find its wavelength. Rearrange the formula to solve for the unknown laser's wavelength \(\lambda_{Unknown}\): \(\lambda_{Unknown} = \frac{2 \Delta d}{\Delta N_{Unknown}}\) Plug in the values to calculate \(\lambda_{Unknown}\): \(\lambda_{Unknown} = \frac{2 * 316.4 * 10^{-6}}{950}\) Calculating, we get: \(\lambda_{Unknown} = 667.2 * 10^{-9}\) meters So, the wavelength of the unknown laser is approximately \(667.2\) nm.

Step by step solution

01

a) Calculate the wavelength of the unknown laser.

First, let's use the formula for the number of fringes formed in an interferometer, which is given by: Number of fringes, \(\Delta N = \frac{2 \Delta d}{\lambda}\) Here, \(\Delta d\) is the distance the movable mirror is displaced, and \(\lambda\) is the wavelength of the laser. We know the reference laser's wavelength and the number of fringes produced, so let's rearrange this formula and find the displacement \(\Delta d\) for the reference laser: \(\Delta d = \frac{\Delta N_{Red} * \lambda_{Red}}{2}\) Plug the given values to calculate \(\Delta d\):

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Most popular questions from this chapter

There is air on both sides of a soap film. What is the smallest thickness of the soap film \((n=1.420)\) that would appear dark if illuminated with \(500 .-\mathrm{nm}\) light?

A \(5.000-\mathrm{cm}\) -wide diffraction grating with 200 grooves is used to resolve two closely spaced lines (a doublet) in a spectrum. The doublet consists of two wavelengths, \(\lambda_{\mathrm{a}}=\) \(629.8 \mathrm{nm}\) and \(\lambda_{\mathrm{b}}=630.2 \mathrm{nm} .\) The light illuminates the entire grating at normal incidence. Calculate to four significant digits the angles \(\theta_{1 \mathrm{a}}\) and \(\theta_{1 \mathrm{~b}}\) with respect to the normal at which the first-order diffracted beams for the two wavelengths, \(\lambda_{\mathrm{a}}\) and \(\lambda_{\mathrm{b}}\), respectively, will be reflected from the grating. Note that this is not \(0^{\circ}\) What order of diffraction is required to resolve these two lines using this grating?

In a single-slit diffraction pattern, there is a bright central maximum surrounded by successively dimmer higher-order maxima. Farther out from the central maximum, eventually no more maxima are observed. Is this because the remaining maxima are too dim? Or is there an upper limit to the number of maxima that can be observed, no matter how good the observer's eyes, for a given slit and light source?

A laser beam with wavelength \(633 \mathrm{nm}\) is split into two beams by a beam splitter. One beam goes to Mirror \(1,\) a distance \(L\) from the beam splitter, and returns to the beam splitter, while the other beam goes to Mirror \(2,\) a distance \(L+\Delta x\) from the beam splitter, and returns to the same beam splitter. The beams then recombine and go to a detector together. If \(L=1.00000 \mathrm{~m}\) and \(\Delta x=1.00 \mathrm{~mm},\) which best describes the kind of interference at the detector? (Hint: To doublecheck your answer, you may need to use a formula that was originally intended for combining two beams in a different geometry, but which still is applicable here.) a) purely constructive b) purely destructive c) mostly constructive d) mostly destructive e) neither constructive nor destructive

A Newton's ring apparatus consists of a convex lens with a large radius of curvature \(R\) placed on a flat glass disc. (a) Show that the distance \(x\) from the center to the air, thickness \(d,\) and the radius of curvature \(R\) are given by \(x^{2}=2 R d\) (b) Show that the radius of nth constructive interference is given by \(x_{\mathrm{n}}=\left[\left(n+\frac{1}{2}\right) \lambda R\right]^{1 / 2} .\) (c) How many bright fringes may be seen if it is viewed by red light of wavelength 700\. \(\mathrm{nm}\) for \(R=10.0 \mathrm{~m},\) and the plane glass disc diameter is \(5.00 \mathrm{~cm} ?\)
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