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

Options

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

Answer

a

Explanation

Path difference, `Delta x = (yd)/(D)`
Here, `y = (5 lambda)/(2)`
and `D = 10 d = 50 lambda ("as" d= 5 lambda)`
So,
`Delta x = ((5 lambda)/(2)) ((5 lambda)/(50 lambda)) = (lambda)/(4)`
Corresponding phase difference will be
`phi = ((2 pi)/(lambda)) (Delta x) = ((2 pi)/(lambda)) ((lambda)/(4)) = (pi)/(2)`
or `(phi)/(2) = (pi)/(4)`
`:. I = I_(0) cos^(2) ((phi)/(2))`
`= I_(0) cos^(2) ((pi)/(4)) = (I_(0))/(2)`