The maximum intensity in Young's double-slit experiment is I₀. Distance between the slits is d = 5λ, where λ 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 = 10d?

Solution

Given,
Maximum intensity = I_o
Distance between the slits, d = 5 $λ$
Distance of screen from the slit, D = 10d

Using the formula,
Path difference, $∆ x = \frac{yd}{D}$
Here,    $y = \frac{d}{2} = \frac{5 λ}{2} (as d = 5 λ) and, D = 10 d = 50 λ$

So,
   $∆ x = (\frac{5 λ}{2}) (\frac{5 λ}{50 λ}) = \frac{λ}{4}$

Corresponding phase difference will be

   $ϕ = (\frac{2 π}{λ}) (∆ x) = (\frac{2 π}{λ}) (\frac{λ}{4})$

$= π / 2$

i.e., $\frac{ϕ}{2} = \frac{π}{4}$

$∴$ Intensity of light, $I = I_{0} \cos^{2} (\frac{ϕ}{2})$

$I = I_{0} \cos^{2} (\frac{π}{4}) = \frac{I_{0}}{2} .$