Question

In Young’s double slit experiment the slits are separated by $0.28 \,mm$ and the screen is placed $1.4\, m$ away. The distance between the central and fourth bright fringe is measured to be $1.2 \,cm$. The wavelength of light used in the experiment is

• A. $6\times 10^{-7}m$
• B. $3\times 10^{-7}m$
• C. $1.5\times 10^{-7}m$
• D. $5\times 10^{-6}m$

Answer 1

Answer: (A)

For constructive interference, $x = n\lambda \frac{D}{d}$
Here, $n =4, d = 0.28\times 10^{-3} m$ and $D = 1.4\,m$
$\therefore \lambda = \frac{xd}{nD} $
$=\frac{ 1.2\times 10^{-2}\times 0.28\times 10^{-3}}{4\times 1.4} $
$= 6\times 10^{-7} m$