The intensity at the maximum in a Young's double slit experiment

The intensity at the maximum in a Young's double slit experiment is I₀. Distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10 d ?

Let P is a point in front of one slit at which intensity is to be calculated. From figure,

Path difference = S₂P – S₁P
= √D² + d² - D

= D		1 +	1		d²		- D
			2		D²

= D		1 +	d²	- 1		=	d²
			2D²				2D

ΔX =	d²
	2 × 10d

	d	=	5λ	=	λ
	20		20		4

Phase difference,

Δφ =	2π	×	λ	=	π
	λ		4		2

So, resultant intensity at the desired point 'p' is

I = I₀ cos²	φ	= I₀ cos²	π	=	I₀
	2		4		2