Chapter 4: Q31E (page 1074)

The microwaves in a certain microwave oven have a wavelength of 12.2 cm. (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing-wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made 5.0 cm longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

Expert verified

Thewidth of the oven containing five antinodal planes is 30.5 cm.
The frequency of these microwaves $2.46 \times 10^{9} Hz$
The frequency of the microwaves when the length of the oven is made 5.0 cm longer is $2.11 \times 10^{9} Hz$

Step by step solution

There is a spacing of half-wavelength between antinodal and nodal planes. In the oven length there is $2 (\frac{1}{2}) λ = L$ wavelengths fitted.

Calculate the length by substituting the required values,

$L = (\frac{5}{2}) (12.2 cm) = 30.5 cm$

Thus , the width of the oven is 30.5 cm.

The frequency is given as,

$f = \frac{c}{λ} = \frac{(3.00 \times 10^{8} m / s)}{(. 122 m)} = 2.46 \times 10^{9} Hz$

Thus, the frequency is $2.46 \times 10^{9} Hz$

In this case L = 35.5 cm given, so, solve for the wavelength,

$λ = \frac{2 L}{5} = \frac{2 (35.5 cm)}{5} = 14.2 cm$

Use this wavelength value to calculate the frequency,

$f = \frac{c}{λ} = \frac{(3.00 \times 10^{8} m / s)}{(0.142 m)} = 2.11 \times 10^{9} Hz$

Thus, the frequency is $2.11 \times 10^{9} Hz$

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