Question

The figure shows a surface $XY$ separating two transparent media, medium-$1$ and medium-$2$. The lines $ab$ and $cd$ represent wave fronts of a light wave traveling in medium-$1$ and incident on $XY$. The lines $ef$ and $gh$ represent wave fronts of the light wave in medium-$2$ after refraction.

The phases of the light wave at c, $d$, e and $f$ are ${\varphi }_c,{\varphi }_d,{\varphi }_e$ and ${\varphi }_f$, respectively. It is given that ${\varphi }_c\ne {\varphi }_f$.

• A. ${\varphi }_c$ cannot be equal to ${\varphi }_d$
• B. ${\varphi }_d$ can be equal to ${\varphi }_e$
• C. $\left({\varphi }_d-{\varphi }_f\right)$ is equal to $\left(f_c-{\varphi }_e\right)$
• D. $\left({\varphi }_d-{\varphi }_c\right)$ is not equal to $\left({\varphi }_f-{\varphi }_e\right)$

Answer 1

Answer: (C)

Optical path length is same for the lines $df$ and $ce$ and hence the phase change is also same:

We have ${\varphi }_d-\varphi )f={\varphi }_c-{\varphi }_e$
as $t_{dQ}+t_{Qf}=t_{cP}+t_{Pe}$.
We know that $Y$ remains unchanged in both media (denser and rarer).
Therefore,
$\gamma t_{dQ}+\gamma t_{Qf}=\gamma t_{cP}+\gamma t_{Pe}$
That is,
$\left(\frac{V_1}{{\lambda }_1}\right)t_{dQ}+\left(\frac{V_2}{{\lambda }_2}\right)t_{Qf}=\left(\frac{V_1}{{\lambda }_1}\right)t_{cP}+\left(\frac{V_2}{{\lambda }_2}\right)t_{Pe}$
$\frac{dQ}{{\lambda }_1}+\frac{Qf}{{\lambda }_2}=\frac{cP}{{\lambda }_1}+\frac{Pe}{{\lambda }_2}$
$\frac{dQ-cP}{{\lambda }_1}=\frac{Pe-Qf}{{\lambda }_2}$
$\frac{2\pi }{{\lambda }_1}(dQ-cP)=\frac{2\pi }{{\lambda }_2}(Pe-Qf)$
$\Rightarrow {\varphi }_d-{\varphi }_c={\varphi }_c-{\varphi }_f(1)$
Since all points on a wave front are at the same phase, we have
${\varphi }_d={\varphi }_c$ and ${\varphi }_f={\varphi }_e(2)$
Here, from (1) and (2),
$\left({\varphi }_d-{\varphi }_f\right)$ is equal to $\left({\varphi }_c-{\varphi }_e\right)$.