Answer to Question #162545 in Optics for joel

Question

Coherent microwaves of wavelength 5.00cm enter a tall, narrow window in a building otherwise essentially opaque to the microwaves. If the window is 36.0cm wide, what is the distance from the central maximum to the first-order minimum along a wall 6.50m from the window?

Answer 1

From the geometry of the problem (sketch of the problem), we can see that distance y from the central maximum to the first-order minimum to the first-order minimum can be found as

$tan\theta = \large\frac{y}{L}$ (1)

where L is the distance from the window (slit) to the wall (screen) and angle $\theta$ is the angle at which we are observing the minimum, which can be calculated using the expression.

$asin\theta = m\lambda$ (2)

In equation (2) a is the width of the window, m = $\pm1, \pm2,...$ is an order of the diffraction (in our case m = 1 because we are interested in the first-order minimum.) and $\lambda$ is wavelength of the wave. Therefore, we can extract the angle $\theta$ from (2).

$\theta = sin^{-1}\large(\frac{\lambda}{a})$ (3)

and inert it to (1) to get y

$y = Ltan\theta$ (Inserting (3))

$y = Ltansin^{-1}\large(\frac{\lambda}{a} )$ = $6.5m*tansin^{-1}\large(\frac{5cm}{36cm})$ = 0.912m = 91.2cm