A ray of light crosses a boundary between two transparent materials. The medium the ray enters has a larger optical density. Whi
1 answer:
Answer:
The wavelength of the light decreases as it enters into the medium with the greater optical density.
The frequency of the light remains constant as it transitions between materials.
Explanation:
- When a ray of light crosses a boundary between two different materials, it undergoes refraction: the ray changes direction, and it also changes speed, according to the relationship:
where v is the speed of the ray of light in the material, c is the speed of light in a vacuum, n is the index of refraction of the material, which is larger for a medium with larger optical density. So, from the equation, we see that the larger the optical density, the smaller the speed of the wave.
- The frequency of the light does not depend on the properties of the medium, so it remains unchanged: therefore the statement
The frequency of the light remains constant as it transitions between materials.
is correct.
- Moreover, the wavelength of the ray of light is related to the speed and the frequency by the equation
where v is the speed and f the frequency. Since we have seen that v decreases and f remains constant, this means that the wavelength decreases as well, so the statement
The wavelength of the light decreases as it enters into the medium with the greater optical density.
is also correct.
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(a) first two overtones for each string:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
while the second overtone (3rd harmonic) is
Similarly, for the second string with fundamental frequency
, the first overtone is
and the second overtone is
(b) The fundamental frequency of a string is given by
where L is the string length, T the tension, and
is the mass per unit of length. This part of the problem says that the tension T and the length L of the string are the same, while the masses are different (let's calle them 
, the mass of the string of frequency 196 Hz, and 
, the mass of the string of frequency 523 Hz.
The ratio between the fundamental frequencies of the two strings is therefore:
and since L and T simplify in the equation, we can find the ratio between the two masses:
(c) Now the tension T and the mass per unit of length
is the same for the strings, while the lengths are different (let's call them 
and 
). Let's write again the ratio between the two fundamental frequencies
And since T and simplify, we get the ratio between the two lengths:
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
and 
. Let's write again the ratio of the frequencies:
Now m and L simplify, and we get the ratio between the two tensions:
The first string has a fundamental frequency of 196 Hz. The n-th overtone corresponds to the (n+1)-th harmonic, which can be found by using
where f1 is the fundamental frequency.
So, the first overtone (2nd harmonic) of the string is
while the second overtone (3rd harmonic) is
Similarly, for the second string with fundamental frequency
and the second overtone is
(b) The fundamental frequency of a string is given by
where L is the string length, T the tension, and
The ratio between the fundamental frequencies of the two strings is therefore:
and since L and T simplify in the equation, we can find the ratio between the two masses:
(c) Now the tension T and the mass per unit of length
And since T and
(d) Now the masses m and the lenghts L are the same, while the tensions are different (let's call them
Now m and L simplify, and we get the ratio between the two tensions: