Question

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

Answer 1

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 $2\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 ${{n}^{th}}$ 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,
${{x}_{n}}=n\dfrac{\lambda D}{d}$
Where
$\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 ${{n}^{th}}$ bright band or a bright fringe is formed when the phase difference between the incoming wave is an integral multiple of $2\pi $ or the path difference between two waves is a full wavelength $\left( \lambda \right)$. If the distance of the screen from the slits is D and the distance between two slits is ‘d’, then the position of the ${{n}^{th}}$ bright band is,
${{x}_{n}}=\dfrac{n\lambda D}{d}$
Where $\lambda $ is the wavelength of light.
The value of n for central bright fringe is zero, so the distance between the ${{n}^{th}}$ bright fringe to the central fringe is,
$\Delta x={{x}_{n}}-{{x}_{0}}$
$\Rightarrow \Delta x=\dfrac{n\lambda D}{d}-\dfrac{(0)\lambda D}{d}$
$\therefore \Delta x=\dfrac{n\lambda D}{d}$
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 $\pi $ or we can say that the path difference between the incident waves is an integral multiple of half the wavelength $\left( \dfrac{\lambda }{2} \right)$. The position of the nth dark band from the central bright band is given by the formula,
${{x}_{n}}=\left( 2n+1 \right)\dfrac{\lambda D}{2d}$
Where
$\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,
$\Delta x={{x}_{d}}-{{x}_{b}}$
$\Rightarrow \Delta x=\dfrac{\left( 2n+1 \right)\lambda D}{2d}-\dfrac{n\lambda D}{d}$
$\therefore \Delta x=\dfrac{n\lambda D}{2d}$