A) \[28.9{}^\circ \]
B) \[26.8{}^\circ \]
C) \[44.5{}^\circ \]
D) \[36.1{}^\circ \]
Correct Answer: C
Solution :
From Eqs. (i) and (ii), we get \[\sin r=\frac{{{\mu }_{1}}}{{{\mu }_{2}}}\sin i\] Given, \[i={{40}^{o}},\,\,{{\mu }_{1}}={{\mu }_{oil}}=1.45\], \[{{\mu }_{2}}={{\mu }_{w}}=1.33\] \[\therefore \] \[\sin r=\frac{1.45}{1.33}\sin {{40}^{o}}\] \[=\frac{1.45}{1.33}\times 0.6428\] \[\Rightarrow \] \[\sin r=0.7007\] \[\Rightarrow \] \[r={{\sin }^{-1}}(0.7007)={{44.5}^{o}}\] Note:- \[\left[ \text{sin 4}0{}^\circ =0.\text{6428},\text{ sin 44}.\text{5}{}^\circ =0.\text{7}00\text{7} \right]\]
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