A) \[\sqrt{\frac{2h}{g}}\]
B) \[\sqrt{\frac{2\ell }{g}}\]
C) \[\frac{1}{\sin \theta }\sqrt{\frac{2h}{g}}\]
D) \[\sin \theta \sqrt{\frac{2h}{g}}\]
Correct Answer: C
Solution :
For TIR at B \[\sin \theta > sin C,~~~~~~~~~~sin\left( 90 - r \right) > sin C.\] \[\cos r > sin C\] .....(1) By snell law \[1\times sini=n\,sin\,r\] \[\sin \,r=\frac{sin\,i}{n}\] \[\cos \,\,r=\sqrt{1-\frac{si{{n}^{2}}\operatorname{i}}{{{\operatorname{n}}^{2}}}}>sin\,C\] \[\sqrt{1-\frac{si{{n}^{2}}\operatorname{i}}{{{\operatorname{n}}^{2}}}}>\frac{1}{\operatorname{n}}\] \[\operatorname{On} solving {{n}^{2}} > si{{n}^{2}} i +1\] \[\operatorname{n} > \sqrt{2} \left( for max value of sin i = 1 \right)\]You need to login to perform this action.
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