Question

Two light rays having the same wavelength $\lambda$ in vacuum are in phase initially. Then the first ray travels a path $L_1$ through a medium of refractive index $n_1$ while the second ray travels a path of length $L_2$ through a medium of refractive index $n_2$. The two waves are then combined to produce interference. The phase difference between the two waves is:

• A. $\frac{2\pi }{\lambda }(L_2-L_1)$
• B. $\frac{2\pi }{\lambda }(n_1{L}_1-n_2{L}_2)$
• C. $\frac{2\pi }{\lambda }(n_2{L}_1-n_1{L}_2)$
• D. $\frac{2\pi }{\lambda }(\frac{L_1}{n_1}-\frac{L_2}{n_2})$

Answer 1

The correct option is B $\frac{2\pi }{\lambda }(n_1{L}_1-n_2{L}_2)$

The optical path between any two points is proportional to the time of travel.
The distance traversed by light in a medium of refractive index $\mu$ in time $t$ is given by
$d=vt$ .....(i)

where $v$ is velocity of light in the medium. The distance traversed by light in a vacuum in this time,

$\Delta =ct$

$=c\cdot \frac{d}{v}$ [from equation (i)]

$=d\frac{c}{v}=\mu d$ .......(ii) (Since, $\mu =\frac{c}{v}$)

This distance is the equivalent distance in vacuum and is called optical path.

Here, optical path for first ray $=n_1{L}_1$

Optical path for second ray $=n_2{L}_2$

Path difference $=n_1{L}_1-n_2{L}_2$

Now, phase difference

$=\frac{2\pi }{\lambda }\times$ path difference

$=\frac{2\pi }{\lambda }\times (n_1{L}_1-n_2{L}_1)$