Question

Two light waves are represented by $y_1=Asin\omega t$ and $y_2=Asin(\omega t+\delta )$ . The phase of the resultant wave is:

• A. $2\delta$
• B. $\frac{\delta }{2}$
• C. $\frac{\delta }{3}$
• D. $\frac{\delta }{4}$

Answer 1

Answer: (B)

$\frac{\delta }{2}$

Resultant wave is:

$y=y_1+y_2$
$=Asinwt+Asin(wt+\delta )$
$=Asinwt+Asinwtcos\delta +Acoswtsin\delta$
$=(A+Acos\delta )sinwt+\left(Asin\delta \right)coswt$
$=\sqrt{(A+Acos\delta {)}^2+(Asin\delta {)}^2}[\frac{A+Acos\delta }{\sqrt{(A+Acos\delta {)}^2+(Asin\delta {)}^2}}sinwt+\frac{Asin\delta }{\sqrt{(A+Acos\delta {)}^2}+(Asin\delta {)}^2}\cos wt]$
Let $\cos {\delta }_1=\frac{A(1+cos\delta )}{\sqrt{A^2(1+cos\delta {)}^2+A^{2}si{n}^2\delta }}$

Then,

$y=\sqrt{(A+A\cos \delta {)}^2}\sin (\omega t+{\delta }_1)$

Hence, phase of the resultant wave is ${\delta }_1$

From above,

$\cos {\delta }_1=\frac{A(1+cos\delta )}{\sqrt{A^2(1+cos\delta {)}^2+A^{2}si{n}^2\delta }}$

$=\frac{A(1+\cos \delta )}{A\sqrt{1+co{s}^2\delta +2cos\delta +{\sin }^2\delta }}$

$=\frac{1+\cos \delta }{\sqrt{2(1+\cos \delta )}}$

$=\frac{\sqrt{1+\cos \delta }}{\sqrt{2}}$

$=\frac{\sqrt{2{\cos }^2\left(\delta /2\right)}}{\sqrt{2}}$

$⟹\delta =\delta /2$