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 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

Explanation

`(sini)/(sinr) =(1)/(n)`
Since, `tanr=(2h)/(2h)=1rArrr=45^@`
`rArr sini=(h)/(hsqrt(5))rArr sini=(1)/(sqrt(5)`
`:. (1)/(n)=(1sqrt(5))/
(1//sqrt(2))rArr n=sqrt((5)/(2))`