A student sees the top edge and the bottom centre \[C\] of a pool simultaneously from an angle \[\theta \] above the horizontal as shown in the figure. |
The refractive index of water which fills up to the top edge of the pool is \[\frac{4}{3}.\operatorname{if}\frac{h}{x}=\frac{7}{4},\] then \[\theta \]is |
A) \[\frac{2}{7}\]
B) \[\frac{8}{3\sqrt{45}}\]
C) \[\frac{8}{3\sqrt{53}}\]
D) \[\frac{8}{21}\]
Correct Answer: C
Solution :
Ray diagram for pool is as shown below. | |
Using \[{{n}_{1}}.\sin i={{n}_{2}}.\sin \,r,\]we have | |
\[1\times \sin (90{}^\circ -\theta )=\frac{4}{3}\sin \,r\] | ? (i) |
Also, \[\tan r=\frac{x}{2h}=\frac{4}{7\times 2}=\frac{2}{7}\]\[\Rightarrow \] \[\sin r=\frac{2}{\sqrt{53}}\] | |
Substituting sin \[r\]in Eq. (i), we have | |
\[\cos \,\theta =\frac{4}{3}\times \frac{2}{\sqrt{53}}=\frac{8}{3\sqrt{53}}\] |
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