Question

For the two parallel rays $\text{(AB and DE)}$ of same wavelength $\lambda$, $\text{BD}$ is the wavefront. The phase difference of these two rays at point $\text{D}$ is

• A. $\frac{\sqrt{2}t\pi }{\lambda }+\pi$
• B. $\frac{\sqrt{5}t\pi }{\lambda }+\pi$
• C. $\frac{2\sqrt{3}t\pi }{\lambda }+\pi$
• D. $\frac{t\pi }{\lambda }+\pi$

Answer 1

The correct option is C $\frac{2\sqrt{3}t\pi }{\lambda }+\pi$


From the figure, we can see the path difference between $\text{AB}$ and $\text{ED}$ is $\text{BC + CD}$.

So, $\Delta x=BC+CD.......\left(1\right)$

On applying concept of geometry,

$\Delta DFC$ and $\Delta DCB$ are congruent from RHS congruent rule.

Therefore, $DF=DB=t$

Now, from trigonometry in $\Delta DBC,$

$BC=\frac{t}{\sqrt{3}}\text{and}CD=\frac{2t}{\sqrt{3}}$

From equation $\left(1\right)$,

$\Delta x=\frac{t}{\sqrt{3}}+\frac{2t}{\sqrt{3}}=\sqrt{3}t$

Further, phase difference,

$\varphi =\frac{2\pi }{\lambda }\times \Delta x=\frac{2\pi }{\lambda }\times \sqrt{3}t$

$\therefore \varphi =\frac{2\sqrt{3}t\pi }{\lambda }$

And due to reflection of ray, ${\varphi }_r=\pi$

$\therefore {\varphi }_{total}=\varphi +{\varphi }_r=\frac{2\sqrt{3}t\pi }{\lambda }+\pi$

Hence, option $\left(\text{C}\right)$ is correct.