Question

In an interference experiment, distance between the slits is 2 mm and screen is placed at distance 1 m from the slits. Fourth dark fringe is formed exactly opposite to one of the slits. Wavelength of light used in nm is

• A. 480
• B. 600
• C. 570
• D. 500

Answer 1

The Correct Option is C

For $n^{th}$ dark fringe, $x_{n} = \left(\frac{2n -1}{2}\right) \frac{\lambda D}{d} $
Here,$ D = 1 m, d=2 \times10^{-3} m , n = 4 $
$\therefore \, x_{4} = \frac{7}{2} \times \frac{\lambda D}{d} $
or, $ \frac{d}{2} = \frac{7}{2} \frac{\lambda D}{d} \hspace10mm \left( \because \, x_{4} = \frac{d}{2}\right) $
$ \therefore \, \, \lambda = \frac{d^{2}}{7D} = \frac{\left(2\times 10^{-3}\right)^{2}}{7\times 1} = \frac{4\times 10^{-6}}{7} $
$ \hspace10mm = 0.571 \times 10^{-6} m $
$\hspace10mm = 571 \times 10^{-9} m $
$\hspace10mm = 570 \: nm $