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givi [52]
3 years ago
8

Rounded to the nearest ten thousand 101,687 and 173,245

Mathematics
1 answer:
Soloha48 [4]3 years ago
6 0
10"1",687 look to the number in the thousands place. If it is 5 or less, do not round up. If it is 5 or more, round up. the "1" is in the thousands place. It is 5 or less, so the answer would be 101,000. 17"3",245 -- the "3" is in the thousands place and is less than 5, so do not round up. = 173,000. 
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Answer:

α = 0.22°

Step-by-step explanation:

To find the angle between the two refracted beams we need to use Snell's Law:

n_{1}sin(\alpha_{1}) = n_{2}sin(\alpha_{2})  

<u>Where:</u>

<em>n₁:</em><em> is the refractive index of the incident beam (air) = 1</em>

<em>n₂:</em><em> is the refractive index of the refractive beam (glass)</em>

<em>α₁:</em><em> is the angle of the incident beam = 60°</em>

<em>α₂:</em><em> is the angle of the refractive beam  </em>

Hence, for the beam with λ = 450 nm and n₂ = 1.4820, the refractive angle (α₂) is:

sin(\alpha_{2}) = \frac{n_{1}}{n_{2}}*sin(\alpha_{1}) = \frac{1}{1.4820}sin(60) = 0.5844                  

\alpha_{2} = 35.76 ^{\circ}

Similarly, the refractive angle for the beam with λ = 700 nm and n₂ = 1.4742 is:

sin(\alpha_{2}) = \frac{n_{1}}{n_{2}}*sin(\alpha_{1}) = \frac{1}{1.4742}sin(60) = 0.5875                  

\alpha_{2} = 35.98 ^{\circ}

Finally, the angle between the two refracted beams is:

\alpha_{2}_{(\lambda=700nm)} - \alpha_{2}_{(\lambda =450nm)} = 35.98 ^{\circ} - 35.76 ^{\circ} = 0.22 ^{\circ}

I hope it helps you!  

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