A plane wave y=a sin (bx+ct) is incident on a surface. Equation of the reflected wave is: y′=a′ sin (ct−bx). Then which of the following statements are correct?
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a
The wave is incident normally on the surface
b
Reflecting surface is y-z plane
c
Medium, in which incident wave is travelling, is denser than the other medium
d
a' cannot be greater than a
answer is A.
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Detailed Solution
If a wave is incident normally on a surface then it gets reflected back to its original path. The incident wave is travelling along negative x-direction and reflected wave is travelling along positive x-direction. Hence, the wave is incident normally on the surface. Therefore, option (1) is correct. The equation of the reflected wave will be y′=a′sin(ct−bx+ϕ) only when the reflecting surface is x = 0 plane, i.e., y - z plane. Hence option (2) is correct.Since, ϕ is equal to zero, it means, no phase change takes place at the reflecting surface. It is possible only when the reflecting surface is boundary of a rarer medium. It means wave is travelling in a denser medium relative to the other medium. Hence, option (3) is also correct. If the reflecting surface is perfectly elastic then whole of the incident energy gets reflected back. In that case a' will be equal to a. But if a part of wave is refracted into the other medium, then the amplitude of oscillations for the reflected wave will be less than that for incident wave. It implies that a' can never be greater than a. Hence, option (4) is also correct.
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A plane wave y=a sin (bx+ct) is incident on a surface. Equation of the reflected wave is: y′=a′ sin (ct−bx). Then which of the following statements are correct?