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Question: The distance of \({{n}^{th}}\) bright band on the screen from the central bright band on either side...

The distance of nth{{n}^{th}} bright band on the screen from the central bright band on either side of the central bright band is _ _ _ _ _ _
A). xn=nλDd{{x}_{n}}=n\dfrac{\lambda D}{d}
B). xn=(2n−1)λDd{{x}_{n}}=\left( 2n-1 \right)\dfrac{\lambda D}{d}
C). xn=nλdD{{x}_{n}}=n\dfrac{\lambda d}{D}
D). xn=(2n−1)λD2d{{x}_{n}}=\left( 2n-1 \right)\dfrac{\lambda D}{2d}

Explanation

Solution

Hint: The bright fringes are formed when the light waves coming at a point on the screen have a phase difference which is an integral multiple of 2π2\pi . At the central bright fringe, the phase difference of the incoming is zero. If we know the position of bright fringe from the central bright fringe. We can find the distance between the nth{{n}^{th}} bright band and the central bright band.
Formula Used:
The position of the nth bright band from the central bright band is given by the formula,
xn=nλDd{{x}_{n}}=n\dfrac{\lambda D}{d}
Where
λ\lambda is the wavelength of light.
D is the distance between the slit and the slits.
d is the separation between the two slits.

Complete step by step answer:
In an interference pattern, the nth{{n}^{th}} bright band or a bright fringe is formed when the phase difference between the incoming wave is an integral multiple of 2π2\pi or the path difference between two waves is a full wavelength (λ)\left( \lambda \right). If the distance of the screen from the slits is D and the distance between two slits is ‘d’, then the position of the nth{{n}^{th}} bright band is,
xn=nλDd{{x}_{n}}=\dfrac{n\lambda D}{d}
Where λ\lambda is the wavelength of light.
The value of n for central bright fringe is zero, so the distance between the nth{{n}^{th}} bright fringe to the central fringe is,
Δx=xn−x0\Delta x={{x}_{n}}-{{x}_{0}}
⇒Δx=nλDd−(0)λDd\Rightarrow \Delta x=\dfrac{n\lambda D}{d}-\dfrac{(0)\lambda D}{d}
∴Δx=nλDd\therefore \Delta x=\dfrac{n\lambda D}{d}
This is true for bright bands in the left side and right side of the central bright band.
So, the answer to the question is option (A)

Note: The condition for a dark band/ fringe in a double-slit interference is that the phase difference is an integral multiple of π\pi or we can say that the path difference between the incident waves is an integral multiple of half the wavelength (λ2)\left( \dfrac{\lambda }{2} \right). The position of the nth dark band from the central bright band is given by the formula,
xn=(2n+1)λD2d{{x}_{n}}=\left( 2n+1 \right)\dfrac{\lambda D}{2d}
Where
λ\lambda is the wavelength of light.
D is the distance between the slit and the slits.
d is the separation between the two slits.
The distance between a nth bright fringe and the nth dark fringe is given by,
Δx=xd−xb\Delta x={{x}_{d}}-{{x}_{b}}
⇒Δx=(2n+1)λD2d−nλDd\Rightarrow \Delta x=\dfrac{\left( 2n+1 \right)\lambda D}{2d}-\dfrac{n\lambda D}{d}
∴Δx=nλD2d\therefore \Delta x=\dfrac{n\lambda D}{2d}