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liberstina [14]
3 years ago
8

Suppose we have a laser emitting a diffraction-limited beam (=632 nm) with a 2-mm diameter. How big a light spot from this lase

r would be produced on the surface of the Moon (distance to the Moon is R=376000 km)? Neglect any effects of the Earth’s atmosphere
Physics
1 answer:
expeople1 [14]3 years ago
5 0

Answer:

Explanation:

λ = wave length = 632 x 10⁻⁹

slit width a = 2 x 10⁻³ m

angular separation of central maxima

= 2 x λ /a

= 2 x 632 x 10⁻⁹ / 2 x 10⁻³

= 632 x 10⁻⁶ rad

width in m of light spot.

= 632 x 10⁻⁶  x 376000 km

= 237.632 km

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According to a periodic table, Krypton was created during the fission of Uranium.

<h3>What is the atomic number?</h3>

<em>Atomic</em> number is a characteristic associated with an element and indicates its number of protons, when a fision occurs, the total number protons is conserved.

Thus, the fission of uranium is led by two elements with <em>atomic</em> numbers 56 and 36. According to a periodic table, those <em>atomic</em> numbers are associated to elements Barium (Ba) and Krypton (Kr), respectively.

According to a periodic table, Krypton was created during the fission of Uranium. \blacksquare

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