Question

In a Young's double slit experiment with slit separation $0.1mm$, one observes a bright fringe at angle $\frac{1}{40}rad$ by using light of wavelength ${\lambda }_1$. When the light of wavelength ${\lambda }_2$ is used a bright fringe is seen at the same angle in the same set up. Given that ${\lambda }_1$ and ${\lambda }_2$ are in visible range $(380nm$ to $740nm)$, their values are

• A. $380nm,500nm$
• B. $625nm,500nm$
• C. $380nm,525nm$
• D. $400nm,500nm$

Answer 1

The correct option is B $625nm,500nm$

Path difference $=d\sin \theta \approx d\theta$
$=0.1\times \frac{1}{40}mm=2500nm$
or bright fringe, path difference must be integral multiple of $\lambda$.
$\therefore 2500=n{\lambda }_1=m{\lambda }_2$
$\therefore {\lambda }_1=625,{\lambda }_2=500$ (from $m=5)$
(for $n=4)$.