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Chapter 25: Problem 73

About how close to each other are two objects on the Moon that can just barely be resolved by the \(5.08-\mathrm{m}-\) (200-in.)-diameter Mount Palomar reflecting telescope? (Use a wavelength of \(520 \mathrm{nm}\).)

Short Answer

Expert verified
Answer: The smallest distance between two objects on the Moon's surface that can just barely be resolved by the 5.08-meter diameter Mount Palomar reflecting telescope, using a wavelength of 520 nm, is approximately 0.0693 km (about 69.3 meters).

Step by step solution

01

Find the angular resolution

First, we need to find the angular resolution of the telescope using the formula: Angular resolution = 1.22 * (wavelength / aperture) Here, the aperture is 5.08 meters, and the wavelength is 520 nm (which we need to convert to meters). To convert nanometers to meters, divide by \(10^9\): 520 nm = \((520 * 10^{-9})\) m Now we can find the angular resolution: Angular resolution = 1.22 * (\((520 * 10^{-9})\) m / 5.08 m)
02

Calculate the angular resolution

Performing the calculation, we get: Angular resolution = 1.22 * (\((520 * 10^{-9})\) m / 5.08 m) = \(1.249 * 10^{-7}\) radians Now we have the angular resolution in radians. To find the distance between two objects on the Moon's surface, we will need to convert the angular resolution to degrees.
03

Convert angular resolution to degrees

To convert radians to degrees, multiply by \(\frac{180}{\pi}\): Angular resolution_degrees = \(\frac{180}{\pi} * 1.249 * 10^{-7}\) radians = \(7.16 * 10^{-6}\) degrees Now we have the angular resolution in degrees.
04

Calculate the linear distance on the Moon's surface

Next, we need to find the smallest distance between two objects on the Moon's surface that can just barely be resolved by the telescope. To do this, we can use the small-angle approximation formula: Distance_between_objects = (Angular_resolution_degrees / 360) * (2 * π * Moon_radius) The average radius of the Moon is approximately 1,737.1 km. Plugging in the values, we get: Distance_between_objects = \((7.16 * 10^{-6}\) degrees / 360) * (2 * π * 1737.1 km)
05

Find the distance between two objects on the Moon's surface

Performing the calculation, we get: Distance_between_objects = \((7.16 * 10^{-6}\) degrees / 360) * (2 * π * 1737.1 km) ≈ 0.0693 km So, two objects on the Moon's surface can just barely be resolved by the \(5.08-\mathrm{m}\)-diameter Mount Palomar reflecting telescope if they are about 0.0693 km (about 69.3 meters) apart.

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Most popular questions from this chapter

Light from a red laser passes through a single slit to form a diffraction pattern on a distant screen. If the width of the slit is increased by a factor of two, what happens to the width of the central maximum on the screen?
Parallel light of wavelength \(\lambda\) strikes a slit of width \(a\) at normal incidence. The light is viewed on a screen that is \(1.0 \mathrm{m}\) past the slits. In each case that follows, sketch the intensity on the screen as a function of \(x\), the distance from the center of the screen, for $0 \leq x \leq 10 \mathrm{cm}$ (a) \(\lambda=10 a\). (b) \(10 \lambda=a,\) (c) \(30 \lambda=a.\)
A double slit is illuminated with monochromatic light of wavelength $600.0 \mathrm{nm} .\( The \)m=0\( and \)m=1\( bright fringes are separated by \)3.0 \mathrm{mm}\( on a screen \)40.0 \mathrm{cm}$ away from the slits. What is the separation between the slits? [Hint: Is the small angle approximation justified?]
A beam of yellow laser light \((590 \mathrm{nm})\) passes through a circular aperture of diameter \(7.0 \mathrm{mm}\). What is the angular width of the central diffraction maximum formed on a screen?
A spectrometer is used to analyze a light source. The screen-to-grating distance is \(50.0 \mathrm{cm}\) and the grating has 5000.0 slits/cm. Spectral lines are observed at the following angles: $12.98^{\circ}, 19.0^{\circ}, 26.7^{\circ}, 40.6^{\circ}, 42.4^{\circ}\( \)63.9^{\circ},\( and \)77.6^{\circ} .$ (a) How many different wavelengths are present in the spectrum of this light source? Find each of the wavelengths. (b) If a different grating with 2000.0 slits/cm were used, how many spectral lines would be seen on the screen on one side of the central maximum? Explain.
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