The intensity at the maximum in a Young's double slit experiment is `I_(0^.)` Distance between two slits is `d=5 lamda,` where `lamda` is the wavelength of light used in the experment. What will be that intensity in front of one of the slit on the screen placed at a distance at a distance D=10 d ?

Options

A. `(3)/(4)I_(0)`
B. `(I_(0))/(2)`
C. `I_(0)`
D. `(I_(0))/(4)`

Answer

B

Explanation

Here, `d=5lambda, D=10d, y=(d)/(2)`
Resultant Intensity at `y=(d)/(2),I_(y)=?`
The path difference between two waves at `y=(d)/(2)`
`Deltax=dtantheta=d xx(y)/(D)=(d xx(d)/(2))/(10d)=(d)/(20)=(5lambda)/(20)=(lambda)/(4)`
Corresponding phase difference, `phi=(2pi)/(lambda)Deltax=(pi)/(2)` .
Now, maximum intensity in Young.s double slit experiment,
`I_("max")=I_(1)+I_(2)+2I_(1)I_(2)`
`I_(0)=4I( :. I_(1)=I_(1)=I)`
`:.I=(I_(0))/(4)`
Required intensity `I_(y)=I_(1)+I_(2)+2I_(1)I_(2)cos.(pi)/(2)=2I=(I_(0))/(2)`