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What is theradius of the illumination when seen above from inside a swimming pool from a depth of 10 m on a sunny day? What is the total angle of view? [ Give refractive index of water is 4/3] |
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Answer» Solution :Give: n = 4/3,d = 10 m Radiua of illumination, R = `(d)/(sqrt(n^(2) - 1))` `R = (10)/(sqrt((4//3)^(2)-1))=(10xx3)/(sqrt(16-9)),R=(30)/(sqrt(7)) = 11.32` To find the ANGLE of the VIEW of the cone, `i_(r) = SIN^(-1) ((1)/(n))` `i_(E)=sin^(-1)((1)/(4//3))=sin^(-1)((3)/(4))=48.6^(@)` The total angle of view is, `2i_(c ) = 2 xx 48.6^(@) = 97.2^(@)` |
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