How far are the goggles from the edge of the pool? - laser goggles
It is night, and they let the glass in a pool that is 5.0 m deep. If you have a laser pointer of 1.5 meters from the pool, the eyes can not illuminate when the laser beam, the water goes from 2.5 m from the edge.
How many drinks from the pool?
Friday, January 1, 2010
Laser Goggles How Far Are The Goggles From The Edge Of The Pool?
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2 comments:
You can use the law of Snell's why. Typical values of the refractive index of air and water are about 1.00 and 1.33, respectively. If n is the refractive index and Snell's Law:
s (sin) ((i) = n (w) sin r)
sin (r) = 0.75sin (i)
Where r is the angle of refraction and the angle of incidence (both relative to the perpendicular to the surface).
The laser beam passes through the water of 2.5 m from the edge, and their goal is 1.5 m above the water line. So:
tan (i) = 2.5/1.5 = 3 / 2 = 1.5
i = 56.3099 ° = 0.9828 rad
sin (i) = sin (0.9828) without (i) = 0.83205
sin (r) = 0.75sin (i) = 0.75 (0.83205) = 0.62404
r = 0.6739 rad = 38.61 °
Be the distance between the edge of the glasses:
d = 2.5 + 5tan (R) = 2.5 + 5 (0.79862) = 2.5 + 3.9931
d = 6.4931 m
You can use the law of Snell's why. Typical values of the refractive index of air and water are about 1.00 and 1.33, respectively. If n is the refractive index and Snell's Law:
s (sin) ((i) = n (w) sin r)
sin (r) = 0.75sin (i)
Where r is the angle of refraction and the angle of incidence (both relative to the perpendicular to the surface).
The laser beam passes through the water of 2.5 m from the edge, and their goal is 1.5 m above the water line. So:
tan (i) = 2.5/1.5 = 3 / 2 = 1.5
i = 56.3099 ° = 0.9828 rad
sin (i) = sin (0.9828) without (i) = 0.83205
sin (r) = 0.75sin (i) = 0.75 (0.83205) = 0.62404
r = 0.6739 rad = 38.61 °
Be the distance between the edge of the glasses:
d = 2.5 + 5tan (R) = 2.5 + 5 (0.79862) = 2.5 + 3.9931
d = 6.4931 m
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