A) 1.2
B) 1.3
C) 1.4
D) 1.5
Correct Answer: D
Solution :
As per the given figure on page 12, \[{{n}_{1}}\sin i={{n}_{r}}\sin r,({{n}_{i}})\sin {{40}^{o}}=(1)\sin {{74.6}^{o}}\] \[{{n}_{1}}=\frac{\sin {{74.6}^{o}}}{\sin {{40}^{o}}}=\frac{0.9617}{0.6428'}{{n}_{1}}=1.5\]You need to login to perform this action.
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