Q. The two slits at a distance of $1\, mm$ are illuminated by the light of wavelength $6.5 \times 10^{-7} m$. The interference fringes are observed on a sereen placed at a distance of $1\, m$. The distance between third dark fringe and fifth bright fringe will be
AMUAMU 2004
Solution:
The fringe width for $n$th bright fringe is given by
$\beta_{n}=n \frac{D \lambda}{d}$
where $d$ is distance between coherent sources, $D$ the distance between screen and source.
For $n =5$
$\beta_{5} =5 \frac{D \lambda}{d}$
For $n$th dark fringe, the fringe width is given by
$\beta_{n}=(2 n-1) \frac{D \lambda}{2 d}$
For third dark fringe
$\beta_{3}=(6-1) \frac{D \lambda}{2 d}=\frac{5 D \lambda}{2 d}$
$\therefore \beta_{5}-\beta_{3} =\frac{5 D \lambda}{d}-\frac{5}{2} \frac{D \lambda}{d}$
$\beta_{5}-\beta_{3} =\frac{5}{2} \frac{D \lambda}{d}$
Given, $\lambda =6.5 \times 10^{-7} m,\,D=1\, m,$
$d =1\, mm =1 \times 10^{-3} m$
$\therefore \beta_{5}-\beta_{3} =\frac{5 \times 1 \times 6.5 \times 10^{-7}}{2 \times 10^{-3}}$
$\beta_{5}-\beta_{3} =1.625 \times 10^{-3} m$
$\beta_{5}-\beta_{3} =1.625\, mm$
$\beta_{5}-\beta_{3} \approx 1.63\, mm$
