I have heard the following statement many times: \"The only manmade object on th

Question

I have heard the following statement many times: "The only manmade object on the surface of the Earth that is visible from outer space is the Great Wall of China". Is this true??? In order to "see" the Wall, our eyes must be able to resolve both sides of the Wall. The maximum width of the Wall is 7.0 m. Let's assume that you are looking down at the Wall from space from a very low-Earth orbit, an altitude of 200 km. The pupil diameter of your eye is 5.0 mm, and its average index of refraction is 1.36. You are looking at the Wall with red light that has a wavelength in vacuum of 630 nm. Neglect the refractive effects of the Earth's atmosphere and calculate the smallest separation distance that you can resolve Earth's surface from this altitude. A) 30.7 m. B) 22.6 m C) 0.03 m D) 41.8 m E) 5.89 m The diffraction pattern from a single slit (width 0.02 mm) is viewed on a screen. The width of the central bright maximum is 2.6 cm. If light with a wavelength of 430 nm is used, what is the distance from the slit to the screen? A) 2.6 cm B) 5.2 cm C) 55.8 cm D) 60.4 cm E) None of these.

Explanation / Answer

17) A)30.7 m

given

h = 200 km = 200*103 m

lamda = 630 nm = 630*10^-9 m

d = 5 mm = 5*10^-3 m

y = ?

we know angular resolution,

theta = 1.22*lamda/d

y/h = 1.22*lamda/d

y = 1.22*lamda*h/d

= 1.22*630*10^-9*200*10^3/(5*10^-3)

= 30.7 m

18) D) 60.4 cm

d = 0.02 mm = 0.02*10^-3 m

lamda = 430 nm

W = 2.6 cm = 0.026 m

let R is the distance from slit to screen

we know, width of central maximum, W = 2*lamda*R/d

==> R = W*d/(2*lamda)

= 0.026*0.02*10^-3/(2*430*10^-9)

= 0.604 m

= 60.4 cm

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