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

Options

A. `(5)/(2)`
B. `sqrt((5)/(2))`
C. `sqrt((3)/(2))`
D. `(3)/(2)`

Answer

B

Explanation

In Fig.
`AC = sqrt((2h^(2)) + h^(2)) = hsqrt(5)`
`BC = sqrt(h^(2) + h^(2)) = hsqrt(2)`
The ray `AC` incident on water surface at `/_r` is refracted along `CD`. If `mu` is refractive index of water, then from Snell's Law,
`mu = (sin (90 - i))/(sin r) = (cos i)/(sin r)`
`= (BD'//BC)/(AE//AC) = (AC)/(BC) = (hsqrt(5))/(sqrt(2))`
`mu = sqrt(5//2)`