Truth table for the given circuit (See figure) is

Options

A. `S_(1)` and `S_(2)` have the same intensities.
B. `S_(1)` and `S_(2)` have a constant phase difference.
C. `S_(1)` and `S_(2)` have the same phase.
D. `S_(1)` and `S_(2)` have the same wavelength.

Answer

A, B, D

Explanation

In Young.s double slit experiment, minimum intensity obtained is, `I_(min)=(sqrt(I_(1))-sqrt(I_(2)))^(2)`
But from figure (b), at the minima,
`I_(min)=0`
`:.0=(sqrt(I_(1))-sqrt(I_(2)))^(2)`
`:.sqrt(I_(1))-sqrt(I_(2))=0`
`:.sqrt(I_(1))=sqrt(I_(2))=`
`I_(1)=I_(2)` Option (A) is correct.
Here interference pattern is stationary. Hence `S_(1)` and `S_(2)` must be coherent sources, phase difference between which remains constant. Hence option (B) is also correct.
Here since `S_(1)` and `S_(2)` are coherent and so light waves emitted from them must have same frequencies and hence same wavelength also. (Because here `c=flamda=` constant `impliesf_(1)lamda_(1)=f_(2)lamda_(2)implies` for `f_(1)=f_(2)` we have `lamda_(1)=lamda_(2)`). Thus option (D) is also correct.
Here phase difference between `S_(1)` and `S_(2)` remains constant. But it is not necessary that this phase difference must be zero only. (i.e. it is not necessary that phases of both the sources must be same only). Hence option (C) is wrong.