A) 1.4
B) 1.3
C) 1.2
D) 1.6
Correct Answer: B
Solution :
[b] Clearly, \[{{i}_{c}}\le 60{}^\circ \] So, maximum possible value of \[{{i}_{c}}\]is \[60{}^\circ \]. Now, \[_{1}{{\mu }_{g}}=\frac{1}{\sin \,{{i}_{C}}}\] \[\frac{{{\mu }_{g}}}{{{\mu }_{i}}}=\frac{1}{{{\operatorname{sini}}_{c}}}\text{ or }{{\mu }_{1}}={{\mu }_{g}}{{\operatorname{sini}}_{c}}=1.5\sin 60{}^\circ =1.5\times \frac{\sqrt{3}}{2}\] \[=1.5\times 0.866=1.299=1.3\]You need to login to perform this action.
You will be redirected in
3 sec