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Two light waves having the same wavelength `lambda` in vacuum are in phase initially. Then the first ray travels a path of length `L_(1)` through a medium of refractive index `mu_(1)`. Then second ray travels a path of length `L_(2)` throug a medium of refractive index `mu_(2)`. The two waves are then combined to observed interference effects. The phase difference between the two, when they interfere, is
A. `(2pi)/(lambda)(mu_(1)L_(1)-mu_(2)L_(2))`
B. `(2pi)/(lambda)(L_(2)-L_(1))`
C. `(2pi)/(lambda)((L_(1))/(mu_(1))-(L_(2))/(mu_(2)))`
D. `(2pi)/(lambda)(mu_(2)L_(1)-mu_(1)L_(2))`

1 Answer

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Best answer
Correct Answer - A
Optical path for 1st ray `=mu_(1)L_(1)`
Optical path for 2nd ray `=mu_(2)L_(2)`
`therefore ` Path difference `=(mu_(1)L_(1)-mu_(2)L_(2))`
Now, phase difference `=(2pi)/(lambda)xx`(path difference)
`=(2pi)/(lambda)(mu_(1)L_(1)-mu_(2)L_(2))`

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