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 $5$th bright fringe at the same point?

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

Answer 1

Answer: (C)

$420nm$

Position of nth bright fringe $y_n=\frac{n\lambda D}{d}$

Given : $\lambda =700nm$

$\therefore$ $y_3=\frac{3\times 700D}{d}$ $⟹y_3=\frac{2100D}{d}$

The 5th bright fringe due to light of wavelength ${\lambda }^{\prime }$ is formed at $y_3$

$\therefore$ $y_5=y_3$

OR $\frac{5\times {\lambda }^{\prime }D}{d}=\frac{2100D}{d}$

$⟹$ ${\lambda }^{\prime }=\frac{2100}{5}=420nm$