Question

1.The displacement of two interfering light waves are y1=4sinwt and y2=3sin(wt+π/2)The amplitude of the resultant wave is a. 5 b. 7 c. 1 d. 0


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Answer 1

resultant amplitude =√4*2+3*2

=√25

=5

Answer 2

Answer:

y = 5 units

Explanation:

It is given that,

Displacement of the first wave, $y_1=4\ sin\omega t$

Displacement of the seconds wave, $y_2=3\ sin(\omega t+\pi/2)$

Here, the two waves are interfering. The resultant amplitude of the wave is given by :

$y=\sqrt{y_1^2+y_2^2}$

$y=\sqrt{4^2+3^2}$

y = 5 units

So, the amplitude of the resultant wave is 5 units. Hence, this is the required solution.