Question

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

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

Answer 1

Answer: (B)

$0.2cm$

Fringe with $\beta =\frac{D\lambda }{d}$

$d=\frac{D\lambda }{\beta }$

$=\frac{6000\times {10}^{-10}\times (40\times {10}^{-2})}{0.012\times {10}^{-2}}$

$=0.2cm$