The maximum intensity of fringes in Young's experiment is I. If one of the slit is closed, then the intensity at that place becomes `I_o`. Which of the following relation is true?

Options

A. `I=I_(0)`
B. `I=2I_(0)`
C. `I=4I_(0)`
D. `I=(I_(0))/(4)`

Answer

C

Explanation

If the two slits are of equal widths, the waves produced by them will have equal intensities I' and I'.
the resultant intensity at any point P on the screen is
`I_(R)=4Icos^(2)phi and ` for `phi=0`, we get
`I_(max)=4I'=1 ` (given) . . . (1)
`I'=I_(0)` (given) . . (2)
from (1) and (2), `I=4I_(0)`