Question

A source of light of wavelength $5000 \,\mathring{A}$ is placed at one end of a table $2\, m$ long and $5 \,mm$ above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is

• A. $2 \times 10^{-5} \: m $
• B. $2 \times 10^{-4} \: m $
• C. $2 \times 10^{-3} \: m $
• D. $2 \times 10^{-2} \: m $

Answer 1

Answer: (B)

Here ,
$\lambda = 5000 \mathring A = 5000 \times 10^{-10} m = 5 \times 10^{-7} m$
$ D - 2 m, d=5 mm = 5 \times 10^{-3} m $
Fringe width,
$\beta = \frac{\lambda D}{d} = \frac{\left(5 \times 10^{-7}m\right)\left(2m\right)}{\left(5\times 10^{-3} m\right)} = 2\times 10^{-4} m $