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What is the answer if there is no outer covering of the pipe? |
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Answer» SOLUTION :If there is no covering, then `i. = sin^(-1) ((1)/(1.68)) approx 36.5^(@),r_(max) = 90^(@) - 36.5^(@) = 53.5^(@)` `"We know" , (sini)/(sinr) = mu` `THEREFORE " " sini = sin53.5^(@) xx 1.65 = 80.3 xx 1.65 = 1.33` `"which is absurd as" sini_(max) = sin90^(@) = 1` `therefore " " r_(max) "MUST be less than" " "53.5^(@) " " "which means" " " 0^(@) le 90^(@)` Thus total internal reflection will taken place for any angle of incidence ranging from `0^(@)` to `90^(@)`. |
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