Question

In Young's double slit experiment, when wavelength used is $6000\,\mathring{A}$ and the screen is $40\, cm$ from the slits, then the fringes are $0.012\, cm$ apart. What is the distance between the slits?

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

Answer 1

Answer: (D)

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