A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 2.60 cm thick flat piece
of crown glass and back to air again. The beam strikes the glass at a 28.0° incident angle.
A) At what angles do the two colors emerge?
B) By what distance are the red and blue separated when they emerge?
A) At what angles do the two colors emerge?
B) By what distance are the red and blue separated when they emerge?
1 answer:
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(a) 2 Hz
The frequency of the nth-harmonic is given by
where
is the fundamental frequency
Therefore, the frequency of the third harmonic of the A () is
while the frequency of the second harmonic of the E () is
So the frequency difference is
(b) 2 Hz
The beat frequency between two harmonics of different frequencies f, f' is given by
In this case, when the strings are properly tuned, we have
- Frequency of the 3rd harmonic of A-note: 1320 Hz
- Frequency of the 2nd harmonic of E-note: 1318 Hz
So, the beat frequency should be (if the strings are properly tuned)
(c) 1324 Hz
The fundamental frequency on a string is proportional to the square root of the tension in the string:
this means that by tightening the string (increasing the tension), will increase the fundamental frequency also*, and therefore will increase also the frequency of the 2nd harmonic.
In this situation, the beat frequency is 4 Hz (four beats per second):
And since the beat frequency is equal to the absolute value of the difference between the 3rd harmonic of the A-note and the 2nd harmonic of the E-note,
and , we have two possible solutions for
:
However, we said that increasing the tension will increase also the frequency of the harmonics (*), therefore the correct frequency in this case will be
1324 Hz