A) \[a=\frac{4b}{3}\]
B) \[c=\frac{5a}{3}\]
C) \[c=-\frac{5a}{3}\]
D) \[a=\frac{-\,4b}{3}\]
Correct Answer: C
Solution :
| Let refracted ray is |
| \[r=a\hat{i}+b\hat{j}+c\hat{k}\] |
| Then, normal to plane of incidence is |
| \[N=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}} \\ 6\sqrt{3} & 8\sqrt{3} & -10 \\ 0 & 0 & 1 \\ \end{matrix} \right|=8\sqrt{3}\hat{i}-6\sqrt{3}\hat{j}\] |
| This must be normal to refracted ray |
| \[\therefore \] \[r\cdot N=0\] |
| \[\Rightarrow \]\[8\sqrt{3}a-6\sqrt{3}b=0\]\[\Rightarrow \]\[b=\frac{4}{3}a\] |
| Also, \[\cos (\pi -\hat{i})=\frac{6\sqrt{3}\hat{i}+8\sqrt{3}\hat{j}-10\hat{k})\hat{k}}{\left| 6\sqrt{3}\widehat{i}+8\sqrt{3}\hat{j}-10\hat{k} \right|\,|\hat{k}|}\] |
| \[=-\frac{1}{2}\] |
| \[\Rightarrow \,\,\,\,\,\,\,\,\,\,\cos (\pi -i)=\cos 120{}^\circ \Rightarrow \,\,\,i=60{}^\circ \] |
| As, \[\sqrt{3}\sin r=\sqrt{2}\sin i\] |
| \[\Rightarrow \]\[\sin r=\frac{1}{\sqrt{2}}\]\[\Rightarrow \]\[r=45{}^\circ \] |
| So, angle between refracted ray and normal \[=45{}^\circ \] |
| \[\cos 45{}^\circ =\frac{(a\hat{i}+b\hat{j}+c\hat{k})\cdot \hat{k}}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\] |
| \[\Rightarrow \] \[\frac{1}{\sqrt{2}}=\frac{c}{\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}}\] |
| \[\Rightarrow \] \[\sqrt{2}c=\sqrt{{{a}^{2}}+{{b}^{2}}+{{c}^{2}}}\] |
| \[\Rightarrow \] \[c=\pm \,\frac{5a}{3}\]\[\Rightarrow \]\[c=-\frac{5a}{3}\] |
| As, positive value is not possible in given case. |
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