Question

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

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

Answer 1

Answer: (B)

Fringe width, $\beta=\frac{d \lambda}{d}$
$d=\frac{D \lambda}{\beta}$
$=\frac{6000 \times 10^{-10} \times\left(40 \times 10^{-2}\right)}{0.012 \times 10^{-2}}$
$=0.2\, cm$