Ahunter is standing on flat ground between two vertical cliffs that are directly opposite one another. he is closer to one cliff than the other. he fires a gun and, after a while, hears three echoes. the second echo arrives 1.05 s after the first, and the third echo arrives 0.827 s after the second. assuming that the speed of sound is 343 m/s and that there are no reflections of sound from the ground, find the distance (in m) between the cliffs.

Answer : distance, d = 463.7 m

Explanation :  

According to the given condition it can be assumed that the first echo would be heard from closest cliff. The second echo is from farther cliff and the third echo is from the reflection between the two cliffs.

Let the distance between the first cliff and the point of firing is x and y is the distance between the second cliff and the point of firing.

Then the first echo will travel 2x distance, second will travel 2y distance and third will travel 2x +2y.

So, using above data :

and

On solving :

x = 321.9 m and y = 141.8 m

Hence, total distance between two cliffs is d = 321.9 m + 141.8 m = 463.7 m

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.