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

`PQ = QR = 2h therefore anglei = 45`
`therefore ST = RT = h = KM = MN`
So , `KS = sqrt(h^(2) + (2h)^(2)) = h = sqrt(5)`
`therefore "sin" r = (h)/(hsqrt(5)) = (1)/(sqrt(5)) therefore mu = ("sini")/("sinr") = ("sin"45^(@))/(1 // sqrt(5)) = sqrt((5)/(2))`