IIT JEE Advanced 2007 Paper 2

Optical path length is same for the lines df and ce and hence the phase change is also same: We have $\phi_d-\phi)f$ = $\phi_c-\phi_e$ as $t_{dQ} + t_{Qf}$ = $t_{cP}+t_{Pe}$. We know that γ remains unchanged in both media (denser and rarer). Therefore, $\gamma t_{dQ}+\gamma t_{Qf}$ = $\gamma t_{cP}+\gamma t_{Pe}$ That is, $\left(\frac{V_1}{\lambda_1}\right)t_{dQ}+\left(\frac{V_2}{\lambda_2}\right)t_{Qf}$ = $\left(\frac{V_1}{\lambda_1}\right)t_{cP}+\left(\frac{V_2}{\lambda_2}\right)t_{Pe}$ $\frac{dQ}{\lambda_1}+\frac{Qf}{\lambda_2}$ = $\frac{cP}{\lambda_1}+\frac{Pe}{\lambda_2}$ $\frac{dQ-cP}{\lambda_1}$ = $\frac{Pe-Qf}{\lambda_2}$ $\frac{2\pi}{\lambda_1} (dQ-cP)$ = $\frac{2\pi}{\lambda_2} (Pe-Qf)$ ⇒ $\phi_d - \phi_c$ = $\phi_c-\phi_f$ (1) Since all points on a wavefront are at the same phase, we have $\phi_d$ = $\phi_c$ and $\phi_f$ = $\phi_e$ (2) Here, from (1) and (2), $(\phi_d-\phi_f)$ is equal to $(\phi_c-\phi_e)$.