Two identical light waves, propagating in the same direction, have a phase difference `delta`. After they superpose, the intensity of the resulting wave will be proportional to

Options

A. `cosdelta`
B. `cos((delta)/(2))`
C. `cos^(2)((delta)/(2))`
D. `cos^(2)delta`

Answer

C

Explanation

For the waves, `y_(1)=A_(1) sin omegat and y_(2)=A_(2) sin (omegat+delta)`
The resultant amplitude is given by
`R^(2)=A_(1)^(2)+A_(2)^(2)+2A_(1)A_(2)cosdelta`
But `A_(1)=A_(2)`
`therefore R^(2)=2A^(2)+2A^(2)cos delta=2A^(2)(1+cosdelta)`
`=2A^(2)[2"cos"^(2)(delta)/(2)]=4A^(2)"cos"^(2)(delta)/(2)`
`therefore R^(2)="cos"^(2)(delta)/(2)`
`therefore I=cos^(2)((delta)/(2))`