A) \[0{}^\circ \]
B) \[12.5{}^\circ \]
C) \[15{}^\circ \]
D) \[22.5{}^\circ \]
Correct Answer: C
Solution :
[c]: For normal incidence, \[i={{0}^{o}},{{r}_{1}}={{0}^{o}}\] As\[{{r}_{1}}+{{r}_{2}}=A\] \[\therefore \]\[{{r}_{2}}=A-{{r}_{1}}={{30}^{o}}\] As\[\mu =\frac{\sin e}{\sin {{r}_{2}}}\] \[\therefore \]\[\sin e=\mu \sin {{r}_{2}}\] \[=\sqrt{2}\sin {{30}^{o}}=\frac{1}{\sqrt{2}}\] \[E={{45}^{o}}\] \[\delta =i+e-A={{0}^{o}}+{{45}^{o}}-{{30}^{o}}={{15}^{o}}\].You need to login to perform this action.
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