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neonofarm [45]
3 years ago
6

Interference occurs with not only light waves but also all frequencies of electromagnetic waves and all other types of waves, su

ch as sound and water waves. Suppose that your physics professor sets up two sound speakers in the front of your classroom and uses an electronic oscillator to produce sound waves of a single frequency. When she turns the oscillator on (take this to be its original setting), you and many students hear a loud tone while other students hear nothing.
The professor adjusts the oscillator to produce sound waves of twice the original frequency. What happens?

a. Some of the students who originally heard a loud tone again hear a loud tone, but others in that group now hear nothing.
b. The students who originally heard a loud tone again hear a loud tone, and the students who originally heard nothing still hear nothing.
c. Among the students who originally heard nothing, some still hear nothing but others now hear a loud tone.
d. The students who originally heard a loud tone now hear nothing, and the students who originally heard nothing now hear a loud tone.

Physics
1 answer:
telo118 [61]3 years ago
4 0

Answer

The answer and procedures of the exercise are attached in the following archives.

Step-by-step explanation:

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

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Answer:

The wavelength of the light is 7200\ \AA.

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

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We need to calculate the wavelength

Using formula of width

\beta=\dfrac{D\lambda}{d}

Put the value into the formula

0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}

\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}

\lambda=7.2\times10^{-7}\ m

\lambda=7200\ \AA

Hence, The wavelength of the light is 7200\ \AA.

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