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. $\dfrac{2\pi }{\lambda }\left( {{L}_{2}}-{{L}_{1}} \right)$
B.$\dfrac{2\pi }{\lambda }\left( {{n}_{1}}{{L}_{1}}-{{n}_{2}}{{L}_{2}} \right)$
C. $\dfrac{2\pi }{\lambda }\left( {{n}_{2}}{{L}_{1}}-{{n}_{1}}{{L}_{2}} \right)$
D. $\dfrac{2\pi }{\lambda }\left( \dfrac{{{L}_{1}}}{{{n}_{1}}}-\dfrac{{{L}_{2}}}{{{n}_{2}}} \right)$

Answer 1

Hint: As a first step find the relation for optical path in terms of the refractive index and distance travelled in that medium. This relation could be found by comparing the distance travelled by light in vacuum (optical path) and that in medium. Now recall the expression for phase difference and substitute the path difference so obtained to find the answer.

Formula used:
Expression for phase difference,
$\phi =\dfrac{2\pi }{\lambda }\Delta x$

Complete step by step solution:
In the question, we are given two light rays that are initially in phase and have the same wavelength$\lambda $. The first ray travels a path ${{L}_{1}}$ in a medium of refractive index${{n}_{1}}$ and the other ray travels ${{L}_{2}}$ in a medium of refractive index ${{n}_{2}}$.We are also said that they combine to produce interference and are asked to find the phase difference between the two.
The distance (d) travelled by light in a medium of refractive index $\mu $ in time t with velocity v is given by,
$d=vt$ ………………………………. (1)
The distance travelled by light in vacuum during the same interval of time,
$x=ct$
Substituting (1),
$x=c\left( \dfrac{d}{v} \right)$
But we know that,
$\mu =\dfrac{c}{v}$
$\Rightarrow x=\mu d$
The distance x is called the optical path.
So, the optical path for the first ray here is,
${{x}_{1}}={{n}_{1}}{{L}_{1}}$
The optical path for second ray is,
${{x}_{2}}={{n}_{2}}{{L}_{2}}$
The path difference$\left( \Delta x \right)$could be given by,
$\Delta x={{n}_{1}}{{L}_{1}}-{{n}_{2}}{{L}_{2}}$ ………………………………….. (2)
Now let us recall the expression for phase difference which is given by,
$\phi =\dfrac{2\pi }{\lambda }\Delta x$
Substituting (2),
$\phi =\dfrac{2\pi }{\lambda }\left( {{n}_{1}}{{L}_{1}}-{{n}_{2}}{{L}_{2}} \right)$
Therefore, we found that the phase difference between the two waves is $\phi =\dfrac{2\pi }{\lambda }\left( {{n}_{1}}{{L}_{1}}-{{n}_{2}}{{L}_{2}} \right)$ .

Hence, option B is the correct answer.

Note:
The factors that affect the optical path are reflection, refraction, dispersion of light and absorption. We could reduce the physical length of any optical device less than the length of the optical path by using the folded optics. Also the phase difference is the interval by which a wave leads or lags by another wave and hence is a relative property.