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.

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