Question

The two slits at a distance of $1mm$ 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 $1m$. The distance between third dark fringe and fifth bright fringe will be

• A. 0.65 cm
• B. 4.8 mm
• C. 1.63 mm
• D. 3.25 cm

Answer 1

Answer: (C)

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
${\beta }_n=(2n-1)\frac{D\lambda }{2d}$
For third dark fringe
${\beta }_3=(6-1)\frac{D\lambda }{2d}=\frac{5D\lambda }{2d}$
$\therefore {\beta }_5-{\beta }_3=\frac{5D\lambda }{d}-\frac{5}{2}\frac{D\lambda }{d}$
${\beta }_5-{\beta }_3=\frac{5}{2}\frac{D\lambda }{d}$
Given, $\lambda =6.5\times 10^{-7}m,D=1m,$
$d=1mm=1\times 10^{-3}m$
$\therefore {\beta }_5-{\beta }_3=\frac{5\times 1\times 6.5\times 10^{-7}}{2\times 10^{-3}}$
${\beta }_5-{\beta }_3=1.625\times 10^{-3}m$
${\beta }_5-{\beta }_3=1.625mm$
${\beta }_5-{\beta }_3\approx 1.63mm$