Question

In an interference experiment, third bright fringe is obtained at a point on the screen with a light of $700nm$. What should be the wavelength of the light source in order to obtain $5th$ bright fringe at the same point?

• A. $500nm$
• B. $630nm$
• C. $750nm$
• D. $420nm$

Answer 1

The correct option is D $420nm$

In an interference experiment,

$X\times \frac{d}{D}=n\lambda$

where, X= position of fringe

d= distance between the two slits

D= distnace of screen from the sits

In the given problem, d, D and X are same for both the cases.

$\therefore n_3{\Lambda }_3=n_5{\Lambda }_5$

$\therefore n_3{\Lambda }_3/n_5={\Lambda }_5$

${\Lambda }_5=\frac{3\times 700}{5}=420$ nm