Question

In Young's double-slit experiment with slit separation $0.1$ mm, one observes a bright fringe at angle $\frac{1}{4^{\circ }}$ rad by using the 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 setup. Given that $1$ and $2$ are in the visible range $(380nm$ to $740nm)$, their values are

• (A) $400nm,500nm$
• (B) $2500nm,1250nm$
• (C) $380nm,525nm$
• (D) $380nm,500nm$

Answer 1

The Correct Option is B

Explanation:
$d\sin \theta =$ integral mutiple of $\lambda$$d\sin \theta =n\lambda$ $\because \theta$ is very small$\therefore \sin \theta =\theta$So,$d\cdot \theta =n\lambda$$0.1mm\times \frac{1}{4}=n\lambda$${10}^{-4}\times 0.25\times {10}^{-1}=n\lambda$$2500\times {10}^9=n\lambda$$\lambda =\frac{2500\times {10}^{-9}}{n}$For $n=1$${\lambda }_1=\frac{2500\times {10}^{-9}}{1}$${\lambda }_1=2500\times {10}^{-9}$${\lambda }_1=2500nm$For $n=2$${\lambda }_2=\frac{2500\times {10}^{-9}}{2}$${\lambda }_2=1250\times {10}^{-9}$$=1250nm$So $d_1=2500nm$${\lambda }_2=1250nm$Hence, the correct option is (B).