You shine a flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon. Assume that the beam leaves the flashlight through a 4.0 −cm aperture, that its white light has an average wavelength of 550 nm, and that the beam spreads due to diffraction only. Express your answer using two significant figures.

12.9 km

Explanation:

θ = 1.22 * (λ/d), where

θ = angle of resolution

λ = wavelength

d = width of slit

Also, the radius of the beam on the moon is

r = lθ, were

l = distance from the earth to the moon, = 384*10^6 m

if we substitute the relation for θ, we should have

r = (1.22 * lλ) / d

Now, we substitute the values into the equation

r = (1.22 * 384*10^6 * 550*10^-9) / 0.04

r = 258 / 0.04

r = 6450 m

Diameter of the beam, d = 2r

d = 2 * 6450

d = 12900 m or 12.9 km

32: from largest to smallest, this hierarchy includes eons, eras, periods, epochs, and ages.

33: answer is c

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scattering of light by tiny particles in air to produce the colors we see on sky, is dependent on the following factors:

(i) wavelength of light

(ii) size of the scattering particle  

(iii) distance traveled by light through the atmosphere.

according to rayleigh's law of scattering, the amount of scattering is inversely proportional to the fourth power of its wavelength. shorter the wavelength, greater is its scattering. hence violet/blue light is scattered more than red.

sun's light is a composite light ( visible light) with wavelengths in the range 400 nm (violet) to 700 nm (red). as sunlight passes through our atmosphere, the molecules of air scatter light according to their wavelengths. during sunrise and sunset, the sun's position is below the horizon. hence light from the sun travels a greater distance in the atmosphere when compared to its position a noon.

as sunlight enters the earth's atmosphere, blue light is scattered away and at the end of its journey in the sky, only red and orange wavelengths remain in the beam.

hence at sunrise and sunset, the sky appears reddish orange.