Question

In Young's double slit experiment, when wavelength used is $600\stackrel{\circ }{A}$ and the screen is $40cm$ from the slits, then the fringes are $0.012cm$ apart. What is the distance between the slits?

• A. $0.024cm$
• B. $2.4cm$
• C. $0.24cm$
• D. $0.2cm$

Answer 1

The correct option is D $0.2cm$

Given, $D=40cm=40\times {10}^{-2}m$
$\lambda =600\stackrel{\circ }{A}=6000\times {10}^{-10}m$
$W=0.012cm=0.012\times {10}^{-2}m$
$\therefore$ Fringe width,
$W=\frac{D\lambda }{d}$
$\Rightarrow d=\frac{D\lambda }{W}$
where, $d=$ distance between the slits
$d=\frac{40\times {10}^{-2}\times 6000\times {60}^{-10}}{0.012\times {10}^{-2}}$
$=2\times {10}^{-3}m=2mm$
$=0.2cm$.