question_answer

An observer can see through a pin-hole the top end of a thin rod of height h, placed as shown in the figure. The beaker height is 3h and its radius h. When the beaker is a filled with a liquid up to a height 2h, he can see the lower end of the rod. Then the refractive index of the liquid is                   

A) \[\frac{5}{2}\]

B) \[\sqrt{\frac{5}{2}}\]

C) \[\sqrt{\frac{3}{2}}\]

D) \[\frac{3}{2}\]

Correct Answer: B

Solution :

[b] For the image of point P to be seen by the observer, it should be formed at point Q. \[\therefore \angle NQS=45{}^\circ \] \[\therefore r=45{}^\circ \] Now in \[\Delta QMA\] \[\angle MQA=45{}^\circ \] \[\therefore MA=QA=h\] \[_{2}^{1}\mu =\frac{\sin r}{\sin i}\] \[=\frac{\sin 45{}^\circ }{\sin i}\]                                               ?.(i) In \[=\frac{\sin 45{}^\circ }{\sin i}\] \[P{{M}^{2}}=4{{h}^{2}}+{{h}^{2}}=5{{h}^{2}}\] \[\therefore \sin i=\frac{h}{\sqrt{5}h}=\frac{1}{\sqrt{5}}\]                                ??.(ii) From (i) and (ii) \[_{2}^{1}\mu =\sqrt{\frac{5}{2}}\]