Interference fringes are produced on a screen by using two light sources of intensities / and 9/. The phase difference between the beams `pi/2` is at point P and `pi`at point Q on the screen. The difference between the resultant intensities at point P and Q is

Options

A. 2I
B. 4I
C. 6I
D. 8I

Answer

C

Explanation

For point P, `I_(1)=I,I_(2)=9I and phi=pi/2`
`because` resultant intensity, `I=I_(1)+I_(2)+2sqrt(I_(1)I_(2)) cosphi`
`therefore I_(P)=I+9I+2sqrt(9xx1)xxcos90^(@)`
`=10I+0=10I`
and for point Q, `I_(1)=I,I_(2)=9I and phi=180^(@)`
and `cos180^(@)=-1`
`therefore I_(Q)=I+9I+2sqrt(9xx1)xxcos180^(@)`
`=101+2xx3(-1)=4I`
`therefore I_(P)-I_(Q)=10I-4I=6I`