Question 390: Waves Optics

Question

In Young's double slit experiment fifth dark fringe is formed opposite to one of the slit. IID is the distance between the slits and the screen and $d$ is the separation between the slits, then the wavelength of light used is

Accepted Answer

In Young's double slit experiment, the position of the $n$th dark fringe is determined by the formula:
$ x_n = \frac{D}{d} (2n-1) \frac{\lambda}{2} $
From this, the wavelength $\lambda$ can be expressed as:
$ \lambda = \frac{2 x_n d}{D (2n-1)} \quad 	ext{.... (i)} $
Given that the fifth dark fringe is formed directly across one of the slits, we have:
$ 2 x_5 = d $
Substituting $n = 5$ into Eq. (i), we find:
$ \lambda = \frac{2 x_5 d}{D (2 	imes 5 - 1)} $
Since $2 x_5 = d$, it follows that:
$ \lambda = \frac{d \cdot d}{D(10-1)} $
Simplifying gives:
$ \lambda = \frac{d^2}{9D} $

Answer

$\frac{d^2}{5 D}$

Answer

$\frac{d^2}{9 D}$

Answer

$\frac{d^2}{6 D}$

Answer

$\frac{d^2}{15 D}$

Waves Optics

In Young's double slit experiment fifth dark fringe is formed opposite to one of the slit. IID is the distance between the slits and the screen and d is the separation between the slits, then the wavelength of light used is

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