Question

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

• A. $\frac{d^{2}}{5D}$
• B. $\frac{d^{2}}{9D}$
• C. $\frac{d^{2}}{6D}$
• D. $\frac{d^{2}}{15D}$

Answer 1

Answer: (B)

Position of $n$ th dark fringes in Young's double slit experiment is given,
$x_{n}=\frac{D}{d}(2 n-1) \frac{\lambda}{2}$
$\Rightarrow \lambda=\frac{2 x_{n} d}{D(2 n-1)}$...(i)
Since, fitth dark fringe is formed opposite to one of the slit,
$\therefore 2 x_{5}=d$ (distance between slits]
Put $n=5$ in Eq. (i), we get
$\lambda=\frac{2 x_{5} d}{D(2 \times 5-1)}$
$=\frac{d \cdot d}{D(10-1)}$ [since, $2 x_{5}=d$]
$=\frac{d^{2}}{9 D}$