Fig. shows a surface XY separating two transparent media, medium 1 and medium 2. Lines ab and cd represent wavefronts of a light wave travelling in medium 1 and incident on XY. Line ef and gh represent wavefront of the light wave in medium 2 after rafraciton.
The phase of the ligth wave at c, d, e, and f are `phi_(c)`, phi_(d), `phi_(e)` and `phi_(f)`, respectively. It is given that `phi_(c ) != phi_(f)`. Then
A. `phi_(c )` cannot be equal to `phi_(d)`
B. `phi_(d )` can be equal to `phi_(d)`
C. `phi_(d ) – phi_(f)` is equal to `phi_(c) – phi_(e)`
D. `phi_(d ) – phi_(c)` is not equal to `phi_(f) – phi_(e)`
The phase of the ligth wave at c, d, e, and f are `phi_(c)`, phi_(d), `phi_(e)` and `phi_(f)`, respectively. It is given that `phi_(c ) != phi_(f)`. Then
A. `phi_(c )` cannot be equal to `phi_(d)`
B. `phi_(d )` can be equal to `phi_(d)`
C. `phi_(d ) – phi_(f)` is equal to `phi_(c) – phi_(e)`
D. `phi_(d ) – phi_(c)` is not equal to `phi_(f) – phi_(e)`
Correct Answer – c
All points on a wavefront are at the same phase
`phi_(d) = phi_(c)` and `phi_(f) = phi_(e)`
`phi_(d) – phi_(f) = phi_(c) – phi_(e)`
Hence, the correct option is (c).