In a stationary wave patternn that forms as a result of reflection of waves from an obstacle , the ratio of the amplitude at an antinode and a node isb = 1.5 . What percentage of the energy passes across the obstacle ?

In a stationary wave patternn that forms as a result of reflection of

1.	In a stationary wave patternn that forms as a result of reflection of waves from an obstacle , the ratio of the amplitude at an antinode and a node isb = 1.5 . What percentage of the energy passes across the obstacle ?
Answer» Solution :As we have STUDIED that when incident WAVE and reflected wave superimpose to produce stationary wave , the ratio of amplitudes at ANTINODE and at node is given by `(A_(MAX))/(A_(min)) = (A_(i) + A_(r))/( A_(i) - A_(r))` This ratio is given as ` 1.5 or 3//2`. `(A_(i) + A_(r))/( A_(i) - A_(r)) = (3)/(2) or (1 + (A_(r))/(A_(i)))/(1 - (A_(r))/(A_(i))) = (3)/(2)` Solving this equation , we get `(A_(r))/(A_(i)) = (1)/(5) rArr (( A_(r))/( A_(i)))^(2) = (1)/(25) or , I_(r) = 0.04 I_(i)` this MEANS `4%` of the incident energy is reflected or `96%` energy passes across the obstacle.