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A cubical vessel with non - transparent walls is so located that eye E of observer cannot see its bottom but can see all of the wall CD. To what height should water be filled in vessel for the observer to see an object O placed at a distance b = 10 cm from the corner C(.^(a)mu_(omega)=(4)/(3)) |
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Answer» 16.5 cm `angle ACB =45^(@)` Let h be the required height NO =h - b = h tan r  or `h = (b)/(1 - tan r)` Now `mu = (SIN 45)/(sin r)` `sin r = (1)/(sqrt(2)r) "" therefore COS r = sqrt(1 - sin^(2) r)` so `h=(b)/(1-(1)/(sqrt(2MU^(2)-1)))=(bsqrt(2mu^(2)-1))/(sqrt(2mu^(2)-1-1))` `therefore h = 26.7 cm` |
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