A triple slit experiment is performed as shown in the figure with monochromaric light of amplitude `a_(0).S_(3)M~~S_(2)M=delta`. The graph between resultant amplitude (A) versus `delta` is plotted. Choose the correct option (s).

Options

A.
B.
C. `(A_(10)+A_(20))/(A_(10)+A_(20))=2`, where `A_(10)`, and `A_(20)` are the values shown in option `A`
D. `(A_(10)+A_(20))/(A_(10)+A_(20))=5`, where `A_(10)` and `A_(20)` are the values shown in option `A`.

Answer

A, C

Explanation

`phi=(2pi)/(lamda)delta`
`y_(2)=a_(0)sin (omega t-kx)`
`impliesy_(1)=a_(0)sin(omegat-kx-phi)` and `y_(3)=a_(0)sin(omegat-kx+phi))`
`A=|a_(0)+2a_(0)cosphi|=a_(0)|(1+2cosphi)|`
`implies A=a_(0)|(1+2cos((2pixd)/(lamdaD))|`
`=a_(0)|(1+2cos((2pidelta)/(lamda))|`

Second Method: Suppose `phi` and `delta` repesent the phase difference and path difference between two connective waves so