Ray Optics

JEE Main level, single correct, medium

106 of 295

Question 106 of 295

Two light waves having the same wavelength λ\lambda in vacuum are in phase initially. Then the first wave travels a path L1{L}_{1} through a medium of refractive index n1{n}_{1}while the second wave travels a path of length L2{L}_{2} through a medium of refractive index n2{n}_{2}. After this the phase difference between the two waves is:

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Answer: (C) 2πλ(n1L1−n2L2)\dfrac{2\pi }{\lambda }\left({n}_{1}{L}_{1}-{n}_{2}{L}_{2}\right)

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 μ\mu in time tt is given by d=vtd=vt .......(i)
Where vv is velocity of light in the medium. The distance traversed by light in a vacuum in this time,
s=cts=ct
=c×dv=c\times \dfrac{d}{v} [from equation (i)]
=dcv=μd=d\dfrac{c}{v}=\mu d .....(ii) (Sinceμ=cv\mu =\dfrac{c}{v})
This distance is the equivalent distance in vacuum and is called optical path.
Here, optical path for first ray =n1L1={n}_{1}{L}_{1}
Optical path for second ray =n2L2={n}_{2}{L}_{2}
Path difference =n1L1−n2L2={n}_{1}{L}_{1}-{n}_{2}{L}_{2}
Now, phase difference =2πλ×=\dfrac{2\pi }{\lambda }\timespath difference
=2πλ×(n1L1−n2L2)=\dfrac{2\pi }{\lambda }\times \left({n}_{1}{L}_{1}-{n}_{2}{L}_{2}\right)

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