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

Answer

A

Explanation

Path difference `Deltax = (yd)/D`
Here, `y = d/2 = (5lambda)/2 (as d = 5 lambda)`
and `D = 10 d = 50 lambda`
Corresponding phase difference will be
`Deltaphi = (2pi)/lambda.Deltax = ((2pi)/lambda) (lambda/4) = pi/2`
or `(Deltaphi)/2 = pi/2`
`therefore I = I_(0) cos^(2) (Deltaphi)/2`
`= I_(0) cos^(2) (pi/4) = I_(0)/2`
Hence, the correct answer is option (1)