1.

An observer can see through a pin-hole the top end of a thin of height h, places as shown in 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 liquidis

Answer»

`(5)/(2)`
`sqrt((5)/(2)`
`sqrt((3)/(2)`
`(3)/(2)`

Solution :`(sini)/(sinr)=(1)/(N)`(i)
Since `tanr=(2h)/(2h)=1`
`rArrr=45^(@)`
`RARR sini=(h)/(hsqrt(5))`
`rArr sini=(1)/(sqrt(5))`
`:. (1)/(n)=(1//sqrt(5))/(1//sqrt(2))rArrn=sqrt((5)/(2))`


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