Question

The resultant intensity of the interference pattern formed by the two waves represented by, Y1=a1cos ωt and Y2=a2cos(π/2-ωt) is: a)a1-a2 b)a1+a2 c)a1^2-a2^2 d a1^2+a2^2

Answer 1

Answer on Question #38155 – Physics – Other

Question: the resultant intensity of the interference pattern formed by the two waves represented by, $y_{1} = a_{1} \cdot \cos(\omega t)$ and $y_{2} = a_{2} \cdot \cos\left(\frac{\pi}{2} - \omega t\right)$ is:

a) $a_1 - a_2$;

b) $a_1 + a_2$;

c) $a_1^2 - a_2^2$;

d) $a_1^2 + a_2^2$.

Solution: let us represent these two waves in the exponential form:

$$y_1 = a_1 \cdot \cos(\omega t) = \operatorname{Re}\left(a_1 \cdot e^{i\omega t}\right),$$ $$y_2 = a_2 \cdot \cos\left(\omega t - \frac{\pi}{2}\right) = \operatorname{Re}\left(a_2 \cdot e^{i\left(\omega t - \frac{\pi}{2}\right)}\right).$$

Here $\operatorname{Re}$ means the real part of the complex number. Using this complex form we can easily obtain the resulting wave:

$$y = y_1 + y_2 = a_1 \cdot e^{i\omega t} + a_2 \cdot e^{i\left(\omega t - \frac{\pi}{2}\right)} = e^{i\omega t} \cdot \left(a_1 + e^{-\frac{i\pi}{2}} a_2\right) = e^{i\omega t} \cdot (a_1 - i a_2).$$

The resultant intensity of the interference pattern is equal to the squared amplitude of the resulting wave:

$$I = |y|^2 = \left| e^{i\omega t} \cdot (a_1 - i a_2) \right|^2 = |(a_1 - i a_2)|^2 = (a_1 - i a_2)(a_1 + i a_2) = a_1^2 + a_2^2$$

Answer: d) $a_1^2 + a_2^2$.