Microwaves of wavelength 5.03 cm are incident on a narrow window that is 30.48 cm wide. If the far wall is 7.27 m away from the window, what is the distance from the central maximum to the first order minimum?

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ANSWER

[tex]122\text{ cm}[/tex]

EXPLANATION

Parameters given:

Wavelength of the microwave, λ = 5.03 cm

Width of the window, a = 30.48 cm

Distance of window from the wall, L = 7.27 m

First, we have to apply the condition for destructive interference using the formula:

[tex]\sin\theta=m\frac{λ}{a}[/tex]

where θ = angle of the first fringe from the central axis.

m = order = 1

Solve for θ:

[tex]\begin{gathered} \sinθ=1*\frac{5.03}{30.48}=0.1650 \\ \\ \theta=\sin^{-1}(0.1650) \\ \\ θ=9.50\degree \end{gathered}[/tex]

Now, find the distance between the central maximum and the first order minimum by applying the formula for the position of the fringe:

[tex]\begin{gathered} \frac{y}{L}=\tanθ \\ \\ y=L\tanθ \end{gathered}[/tex]

Solve for y:

[tex]\begin{gathered} y=7.27*\tan9.5 \\ \\ y=1.22\text{ }m=122\text{ }cm \end{gathered}[/tex]

That is the distance from the central maximum to the first order minimum.