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ryzh [129]
3 years ago
14

How long does it take (in minutes) for light to reach venus from the sun, a distance of 1.152 × 108 km?

Physics
1 answer:
7nadin3 [17]3 years ago
7 0
Using the precise speed of light in a vacuum (299,792,458 \ \frac{m}{s}), and your given distance of 1.152 * 10^{8} km, we can convert and cancel units to find the answer. The distance in m, using \frac{1000 \ m}{1 \ km}, is 1.152 * 10^{11} m. Next, for the speed of light, we convert from s to min, using \frac{1 \ min}{60 \ s}, so we divide the speed of light by 60. Finally, dividing the distance between the Sun and Venus by the speed of light in km per min, we find that it is 6.405 min.

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Estimate how far apart the rays of deepest red and deepest violet light are as they exit the bottom surface. assume nred = 1.57
Harlamova29_29 [7]
We begin by noting that the angle of incidence is the one that's taken with respect to the normal to the surface in question. In this case the angle of incidence is 30. The material is Flint Glass according to the original question. The refractive indez of air n1=1, the refractive index of red in flint glass is nred=1.57, finally for violet in the glass medium is nviolet=1.60. Snell's Law dictates:
n_1sin(\theta_1)=n_2sin(\theta_2)
Where \theta_2 differs for each wavelenght, that means violet and red will have different refractive indices in the glass.
In the second figure provided details are given on which are the angles in question, \Delta x is the distance between both rays.
\theta_{2red}=Asin(\frac{sin(30)}{1.57})\approx 18.5705
\theta_{2violet}=Asin(\frac{sin(30)}{1.60})\approx 18.21
At what distance d from the incidence normal will the beams land at the bottom?
For violet we have:
d_{violet}=h.tan(\theta_{2violet})\approx 0.0132m
For red we have:
d_{red}=h.tan(\theta_{2red})\approx 0.0134m
We finally have:
\Delta x=d_{red}-d_{violet}\approx2.8\times10^{-4}m


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