A source S and a detector D are separated by a distance d . As shown in the figure, a large cardboard is positioned at a distance 2 d from the source and the detector. The source emits a wave with a wavelength of d/2

Question

Quantrex Academy · Accepted Answer

Correct answer: B. 0.13 d

Path length of the direct wave from S to D is x_1 = d . Path length of the reflected wave is x_2 = 2 ( 2 d)^2 + ((d)/(2))^2 = 2 2d^2 + (d^2)/(4) = 3d . Initial path difference is Δ x = x_2 - x_1 = 3d - d = 2d . Given wavelength λ = (d)/(2) , the path difference is Δ x = 4λ . This confirms the waves are initially in phase. To make the waves out of phase by shifting the cardboard away by a minimum distance, the new path difference must be Δ x' = 4.5λ = 4.5 ((d)/(2)) = 2.25d . Let the new distance of the cardboard be y . The new path length of the reflected wave is x_2' = 2 y^2 + ((d)/(2))^2 . The new path difference is: x_2' - x_1 = 2.25d 2 y^2 + (d^2)/(4) - d = 2.25d 2 y^2 + (d^2)/(4) = 3.25d y^2 + (d^2)/(4) = 1.625d Squaring both sides: y^2 + 0.25d^2 = (1.625)^2 d^2 = 2.640625 d^2 y^2 = 2.390625 d^2 y = 2.390625 d 1.546 d The minimum shift required is: Δ y = y - 2 d 1.546 d - 1.414 d = 0.132 d 0.13 d . Answer: 0.13 d

A source S and a detector D are separated by a distance d . As...

Options

Correct answer

Step-by-step solution

More from Sound Waves