The maximum intensity in Young's double slit experiment is $I_{0}$ . Distance between the slits is $d = 5 λ$ , 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 $D = 10 d$ ?

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Answer 1

Path difference $= Δ x = \frac{x d}{D}$ -------(i)
In front of one of the slits,
$x = \frac{d}{2}$ but $d = 5 λ$
$x = \frac{5 λ}{2}$ and $D = 10 d = 50 λ$
Now, from equation (i), we have:
$Δ x = \frac{5 λ}{2} \times \frac{5 λ}{5 0 λ} = \frac{λ}{4}$
So, corresponding phas defference:
$ϕ = \frac{2 π}{λ} . (Δ x) = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2}$
As, $I = I_{0} cos^{2} (\frac{ϕ}{2})$
So, $I = I_{0} cos^{2} (\frac{π}{4}) = \frac{I _{0}}{2}$

Answer 2

Path difference

= Δ x = \frac{x d}{D}

-------(i)

In front of one of the slits,

x = \frac{d}{2}

but

d = 5 λ

x = \frac{5 λ}{2}

and

D = 10 d = 50 λ

Now, from equation (i), we have:

Δ x = \frac{5 λ}{2} \times \frac{5 λ}{5 0 λ} = \frac{λ}{4}

So, corresponding phas defference:

ϕ = \frac{2 π}{λ} . (Δ x) = \frac{2 π}{λ} \times \frac{λ}{4} = \frac{π}{2}

As,

I = I_{0} cos^{2} (\frac{ϕ}{2})

So,

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