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What are the approximate dimensions of the smallest object on Earth that astronauts can resolve by eye when they are orbiting \(250 \mathrm{km}\) above the Earth? Assume that \(\lambda=500 \mathrm{nm}\) and that a pupil diameter is \(5.00 \mathrm{mm}\)

Short Answer

Expert verified
The smallest object on Earth that astronauts can see with their eyes when they are orbiting 250 km above the Earth is approximately 30.3 meters.

Step by step solution

01

Convert all measurements to the same units

Before inserting the given values, make sure they are all in the same units. Convert the diameter of the pupil from millimeters to meters: \( D = 5.00 \mathrm{mm} = 5.00 \times 10^{-3} \mathrm{m} \). Convert the wavelength from nanometers to meters: \( \lambda = 500 \mathrm{nm} = 500 \times 10^{-9} \mathrm{m} \).
02

Use the Rayleigh criterion to find the angular resolution

Substitute the values of \( \lambda \) and \( D \) into the Rayleigh criterion formula: \( \Theta = 1.22 (\lambda / D) = 1.22 ((500 \times 10^{-9}) / (5.00 \times 10^{-3})) = 1.22 \times 10^{-4} \mathrm{radians} \).
03

Convert the angle from radians to degrees

It's generally easier to understand angles in degrees, so convert \( \Theta \) from radians to degrees using the conversion \( 1 \mathrm{radian} = 180^\circ/\pi \). Then, \( \Theta = 1.22 \times 10^{-4} \mathrm{radians} * (180/\pi) = 0.00698^\circ \)
04

Calculate the resolution limit on Earth

Since we are looking for the smallest object visible on Earth, we can use the definition of the tangent of an angle in a right triangle: \( \tan(\Theta) = Opposite/List of vertebrates = h / 250 \mathrm{km} \). The height \( h \) is what we are solving for. So, \( h = \tan(\Theta) * 250 \mathrm{km} = \tan(0.00698^\circ) * 250000 \mathrm{m} = 30.3 \mathrm{m} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Rayleigh Criterion
Understanding the Rayleigh criterion is essential in the field of optics, especially when it comes to astronomical observations. In simple terms, the Rayleigh criterion gives us a way to determine the minimum angular separation at which two point light sources (like stars or distant satellites) can be perceived as separate entities through an optical instrument. According to this criterion, two sources are resolvable when the principal diffraction maxima of one image coincides with the first diffraction minima of the other.

The formula that expresses the Rayleigh criterion is:\[ \Theta = 1.22 \frac{\lambda}{D} \]where \( \Theta \) is the angular resolution in radians, \( \lambda \) is the wavelength of light, and \( D \) is the diameter of the aperture (like a telescope's primary mirror or a camera's lens). The factor 1.22 comes from the physics of diffraction through a circular aperture. The better the angular resolution, the finer the details that can be distinguished in the astronomical images captured through telescopes.

Improving angular resolution in telescopes may involve using larger aperture diameters or observing at shorter wavelengths. Both methods would result in a smaller value of \( \Theta \), indicating higher resolution. Despite technological advances, the Rayleigh criterion still sets fundamental limits on what can be resolved, whether with the human eye or sophisticated telescopes.
Angular Resolution Calculation
To calculate angular resolution, astronomers apply the fundamentals of the Rayleigh criterion. By working through the steps of the Rayleigh criterion, one can determine the smallest angle of separation that an optical system can resolve. The calculation process requires understanding both the wavelength of light being observed and the size of the aperture used.

For instance, if an astronaut wishes to determine the smallest object visible on Earth while orbiting at 250 km altitude, the formula derived from the Rayleigh criterion can be used. The angular resolution \( \Theta \), calculated through the equation \( \Theta = 1.22 \frac{\lambda}{D} \), depends directly on the physical properties of light and the optical system. By following this process, the astronaut can identify how fine the details are that they can distinguish from their unique vantage point in space.

Using the provided data, converting units appropriately, and applying trigonometry, the astronaut can find out that the smallest resolvable object on Earth's surface is approximately 30.3 meters wide. This calculation is invaluable for various applications, including Earth observation and surveillance from space, as well as for enhancing our understanding of optics in general.
Astronomical Observation
Astronomical observation is a fascinating field that encompasses studying celestial objects like stars, planets, and galaxies. It's a window to the universe and relies heavily on the ability of telescopes and other observing instruments to resolve fine details in the night sky. Angular resolution plays a crucial role in these observations, defining the capability of a telescope to distinguish between closely spaced celestial bodies.

When astronomers peer into space, they are often limited by the angular resolution of their instruments. The Earth's atmosphere can also blur images, which is why space telescopes like the Hubble Space Telescope, are so valuable. They orbit above the Earth's distorting atmosphere, providing clearer and more detailed views of the universe.

To further enhance the quality of astronomical observation, techniques such as adaptive optics are used to correct atmospheric distortion, and interferometry allows for the combination of information from multiple telescopes to simulate a larger aperture. These advancements aid in pushing the boundaries of the Rayleigh criterion, providing even better resolutions and deeper insights into the cosmos.

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Most popular questions from this chapter

An American standard television picture is composed of about 485 horizontal lines of varying light intensity. Assume that your ability to resolve the lines is limited only by the Rayleigh criterion and that the pupils of your eyes are \(5.00 \mathrm{mm}\) in diameter. Calculate the ratio of minimum viewing distance to the vertical dimension of the picture such that you will not be able to resolve the lines. Assume that the average wavelength of the light coming from the screen is \(550 \mathrm{nm}\)

A helium-neon laser \((\lambda=632.8 \mathrm{nm})\) is used to calibrate a diffraction grating. If the first-order maximum occurs at \(20.5^{\circ},\) what is the spacing between adjacent grooves in the grating?

A wide beam of laser light with a wavelength of \(632.8 \mathrm{nm}\) is directed through several narrow parallel slits, separated by \(1.20 \mathrm{mm},\) and falls on a sheet of photographic film \(1.40 \mathrm{m}\) away. The exposure time is chosen so that the film stays unexposed everywhere except at the central region of each bright fringe. (a) Find the distance between these interference maxima. The film is printed as a transparency-it is opaque everywhere except at the exposed lines. Next, the same beam of laser light is directed through the transparency and allowed to fall on a screen \(1.40 \mathrm{m}\) beyond. (b) Argue that several narrow parallel bright regions, separated by \(1.20 \mathrm{mm},\) will appear on the screen, as real images of the original slits. If at last the screen is removed, light will diverge from the images of the original slits with the same reconstructed wave fronts as the original slits produced. (Suggestion: You may find it useful to draw diagrams similar to Figure \(38.16 .\) A train of thought like this, at a soccer game, led Dennis Gabor to the invention of holography.)

Light from an argon laser strikes a diffraction grating that has 5310 grooves per centimeter. The central and firstorder principal maxima are separated by \(0.488 \mathrm{m}\) on a wall \(1.72 \mathrm{m}\) from the grating. Determine the wavelength of the laser light.

Two closely spaced wavelengths of light are incident on a diffraction grating. (a) Starting with Equation 38.10, show that the angular dispersion of the grating is given by \(\frac{d \theta}{d \lambda}=\frac{m}{d \cos \theta}\) (b) A square grating \(2.00 \mathrm{cm}\) on each side containing 8000 equally spaced slits is used to analyze the spectrum of mercury. Two closely spaced lines emitted by this element have wavelengths of \(579.065 \mathrm{nm}\) and \(576.959 \mathrm{nm} .\) What is the angular separation of these two wavelengths in the second-order spectrum?

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