PHY 101 4th Assignment idea solution..
For question no 1 :- Find Angle Refraction ???
look at this formula
Refraction is described by Snell's law, which states that for a given pair of media and a wave with a single frequency, the ratio of the sines of the angle of incidence θ1 and angle of refraction θ2 is equivalent to the ratio of phase velocities (v1 / v2) in the two media, or equivalently, to the opposite ratio of the indices of refraction (n2 / n1):
so from there
sinθ2=n2 (sinθ1/n1)
n1=1.00
there n2=1.52
θ1=30degree
For question no 2 :-
A radio transmitting station operating at a frequency of 120MHz has two identical
antennas that radiate in phase. Antenna B is 9.00m to the right of A. Consider point P
between the antennas and along the line connecting them, a horizontal distance x to the
right of antenna A. For what values of x will constructive interference occur at point P?
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Solution:-
As far i think .... look at this example please.... where the same condition is given.
Two loudspeakers, A and B are driven by the same amplifier and
emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A.
The frequency of the sound waves produced by the loudspeakers is 206 Hz.
Consider point P between the speakers and along the line connecting them,
a distance x to the right of speaker A. Both speakers emit sound waves
that travel directly from the speaker to point P.
b) For what values of x will constructive interference occur at point P?
----
Solution of example:-
Assume speed of sound c = 343 m/s, then wavelength lambda = c/f = 1.665 m and lambda/4 = 0.41626 m. There are two locations between the speakers where there is interference. At the center the path lengths are equal.
For constructive interference you want a pathlength difference equal to lambda multiplied by an integer>=0. Then you have one constructive point at the center and, using similar logic as in A, two more at lambda/2 = 0.83252 m to either side. No more points can fit in that space. The locations are at 1.03 - 0.83252 and 1.03 + 0.83252 m.
----
f (frequency) = v (velocity)/ L
v=f*L
L=9.00m
f=120
v=120*(9)=1080
For question no 1 :- Find Angle Refraction ???
look at this formula
Refraction is described by Snell's law, which states that for a given pair of media and a wave with a single frequency, the ratio of the sines of the angle of incidence θ1 and angle of refraction θ2 is equivalent to the ratio of phase velocities (v1 / v2) in the two media, or equivalently, to the opposite ratio of the indices of refraction (n2 / n1):
so from there
sinθ2=n2 (sinθ1/n1)
n1=1.00
there n2=1.52
θ1=30degree
For question no 2 :-
A radio transmitting station operating at a frequency of 120MHz has two identical
antennas that radiate in phase. Antenna B is 9.00m to the right of A. Consider point P
between the antennas and along the line connecting them, a horizontal distance x to the
right of antenna A. For what values of x will constructive interference occur at point P?
---------
Solution:-
As far i think .... look at this example please.... where the same condition is given.
Two loudspeakers, A and B are driven by the same amplifier and
emit sinusoidal waves in phase. Speaker B is 2.00 m to the right of speaker A.
The frequency of the sound waves produced by the loudspeakers is 206 Hz.
Consider point P between the speakers and along the line connecting them,
a distance x to the right of speaker A. Both speakers emit sound waves
that travel directly from the speaker to point P.
b) For what values of x will constructive interference occur at point P?
----
Solution of example:-
Assume speed of sound c = 343 m/s, then wavelength lambda = c/f = 1.665 m and lambda/4 = 0.41626 m. There are two locations between the speakers where there is interference. At the center the path lengths are equal.
For constructive interference you want a pathlength difference equal to lambda multiplied by an integer>=0. Then you have one constructive point at the center and, using similar logic as in A, two more at lambda/2 = 0.83252 m to either side. No more points can fit in that space. The locations are at 1.03 - 0.83252 and 1.03 + 0.83252 m.
----
f (frequency) = v (velocity)/ L
v=f*L
L=9.00m
f=120
v=120*(9)=1080
