Question 361: Waves Optics

Question

In Young's double slit experiment, the distance of $$\mathrm{n}^{	ext {th }}$$ dark band from the central bright band in terms of bandwidth '$$\beta$$' is

Accepted Answer

In Young's double slit experiment, the position of dark and bright fringes on the screen is a result of constructive and destructive interference of light waves coming from the two slits.
The distance between two consecutive bright or dark fringes is called the fringe width, denoted by $$\beta$$. The fringe width is given by:
$$ \beta = \dfrac{\lambda D}{d} $$
where:
$$\lambda$$ is the wavelength of light used.
$$D$$ is the distance between the slits and the screen.
$$d$$ is the separation between the two slits.
The position of the nth dark band from the central bright band is given by the condition for destructive interference, which occurs at:
$$ y_n = \left(n - \dfrac{1}{2}ight) \beta $$
where $$n$$ is the order of the dark fringe.
Therefore, the distance of the $$\mathrm{n}^{	ext{th}}$$ dark band from the central bright band is:
Option C: $$(\mathrm{n}-0.5) \beta$$

Answer

$$\mathrm{n} \beta$$

Answer

$$(\mathrm{n}-1) \beta$$

Answer

$$(\mathrm{n}-0.5) \beta$$

Answer

$$(\mathrm{n}+0.5) \beta$$

Waves Optics

In Young's double slit experiment, the distance of $\mathrm{n}^{\text {th }} dark band from the central bright band in terms of bandwidth '\beta$' is

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