A) \[{{n}_{1}}={{n}_{2}}={{n}_{3}}\]
B) \[{{n}_{1}}={{n}_{2}}\ne {{n}_{3}}\]
C) \[1+{{n}_{1}}={{n}_{2}}+{{n}_{3}}\]
D) \[1+n_{2}^{2}=n_{1}^{2}+n_{3}^{2}\]
Correct Answer: D
Solution :
At B \[\sin i={{n}_{1}}\sin {{r}_{1}}\] Þ \[{{\sin }^{2}}i=n_{1}^{2}{{\sin }^{2}}{{r}_{1}}\] .... (i) At C \[{{n}_{1}}\sin (90-{{r}_{1}})={{n}_{2}}\sin {{r}_{2}}\]Þ\[n_{2}^{2}{{\sin }^{2}}{{r}_{2}}=n_{1}^{2}{{\cos }^{2}}{{r}_{1}}\]....(ii) At D \[{{n}_{2}}\sin (90-{{r}_{2}})={{n}_{3}}\sin {{r}_{3}}\]Þ\[n_{2}^{2}{{\cos }^{2}}{{r}_{2}}=n_{3}^{2}{{\sin }^{2}}{{r}_{3}}\] ....(iii) At E \[{{n}_{3}}\sin (90-{{r}_{3}})=(1)\sin (90-1)\]Þ\[{{\cos }^{2}}i=n_{3}^{2}{{\cos }^{2}}{{r}_{3}}\] ....(iv) Adding (i), (ii), (iii) and (iv) we get \[1+n_{2}^{2}=n_{1}^{2}+n_{3}^{2}\]You need to login to perform this action.
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