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Chapter 34: Problem 43

Calculate and compare the angular resolutions of the Hubble Space Telescope (aperture diameter \(2.40 \mathrm{~m}\), wavelength \(450 . \mathrm{nm}\); illustrated in the text), the Keck Telescope (aperture diameter \(10.0 \mathrm{~m}\), wavelength \(450 . \mathrm{nm}\) ), and the Arecibo radio telescope (aperture diameter \(305 \mathrm{~m}\), wavelength \(0.210 \mathrm{~m}\) ). Assume that the resolution of each instrument is diffraction limited.

Short Answer

Expert verified
Answer: The order of image quality from highest to lowest is: Keck Telescope, Hubble Space Telescope, and Arecibo Radio Telescope.

Step by step solution

01

Hubble Space Telescope Resolution

To calculate the angular resolution of the Hubble Space Telescope, we will use the given values for the aperture diameter (D) as \(2.40 \mathrm{m}\) and the wavelength (λ) as \(450 \mathrm{nm}\). Plug these values into the formula and convert the wavelength to meters for consistency: $$ \theta_{Hubble} = 1.22\frac{450 \times 10^{-9} m}{2.4 m} $$
02

Keck Telescope Resolution

To calculate the angular resolution of the Keck Telescope, we will use the given values for the aperture diameter (D) as \(10.0 \mathrm{m}\) and the wavelength (λ) as \(450 \mathrm{nm}\). Plug these values into the formula and convert the wavelength to meters for consistency: $$ \theta_{Keck} = 1.22\frac{450 \times 10^{-9} m}{10.0 m} $$
03

Arecibo Radio Telescope Resolution

To calculate the angular resolution of the Arecibo Radio Telescope, we will use the given values for the aperture diameter (D) as \(305 \mathrm{m}\) and the wavelength (λ) as \(0.210 \mathrm{m}\). Plug these values into the formula: $$ \theta_{Arecibo} = 1.22\frac{0.210 m}{305 m} $$
04

Calculate The Resolutions

Now, calculate the angular resolutions for each telescope: $$ \theta_{Hubble} = 2.275 \times 10^{-7} radians \\ \theta_{Keck} = 5.46 \times 10^{-8} radians \\ \theta_{Arecibo} = 8.524 \times 10^{-4} radians $$
05

Compare The Resolutions

Now that we have the angular resolutions for each telescope, we can compare them. Lower angular resolution means better image quality. From the calculated values, we can see that the order of image quality (from highest to lowest) is: Keck Telescope, Hubble Space Telescope, and Arecibo Radio Telescope.

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Most popular questions from this chapter

A diffraction grating has \(4.00 \cdot 10^{3}\) lines \(/ \mathrm{cm}\) and has white light \((400 .-700 . \mathrm{nm})\) incident on it. What wavelength(s) will be visible at \(45.0^{\circ} ?\)

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

When two light waves, both with wavelength \(\lambda\) and amplitude \(A\), interfere constructively, they produce a light wave of the same wavelength but with amplitude \(2 A .\) What will be the intensity of this light wave? a) same intensity as before b) double the intensity c) quadruple the intensity d) not enough information

For a double-slit experiment, two 1.50 -mm wide slits are separated by a distance of \(1.00 \mathrm{~mm} .\) The slits are illuminated by a laser beam with wavelength \(633 \mathrm{nm} .\) If a screen is placed \(5.00 \mathrm{~m}\) away from the slits, determine the separation of the bright fringes on the screen.

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.
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