The maximum intensity in young's double-slit experiment is `I_(0)`. Distance between the slit is `d = 5 lambda`, where `lambda` is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance `D = 10 d`?

Options

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

Answer

C

Explanation

Here, `d =5 lambda. I = ? D = 10 d, x = d//2`
Path diff. `(x d)/(D) = (d)/(2).(5 lambda)/((10 d)) = (lambda)/(4)`
Phase diff. `= phi = (2 pi)/(lambda) xx (lambda)/(4) = (pi)/(2) = 90^(@)`
`I = I_(0) "cos"^(2) (phi)/(2) = I_(0)("cos"(pi)/(4))^(2) = I_(0) xx ((1)/(sqrt(2)))^(2) = (I_(0))/(2)`