Sunday, 1 July 2012

Phy101 Assignment Solution spring 2012


Question # 1
What is the angle of refraction here, shown in above fig?
Solution:-
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
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:-
    



Phy101 Assignment Solution spring 2012

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.
well wisher at http://www.vucybarien.com
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
you can take hint from this example
just hint not perfect solution …….
question no 3
How is it possible to determine the direction of the polarizing axis of a single polarizer?
Solution:-
According to my opinion the vibration move in one direction, by that we can determine
the direction of the polarizing axis of a single polarize.
Remember me in your prays …..!!!!!!!
MEHRAN ALI SHAH
VIBD01

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