Q. In Young’s double slit experiment, the fringe width is $1 \times 10^{- 4}m$. If the distance between the slit and screen is doubled and the distance between the two slit is reduced to half and wavelength is changed from $6.4 \times 10^{- 7}m$ to $4.0 \times 10^{- 7}m$, the value of new fringe width will be
A) $0.15 \times 10^{- 4}m$
B) $2.0 \times 10^{- 4}m$
C) $1.25 \times 10^{- 4}m$
D) $2.5 \times 10^{- 4}m$
Explanation
(d) $\beta = \frac{\lambda D}{d} \Rightarrow \frac{\beta_{2}}{\beta_{1}} = \frac{\lambda_{2}D_{2}d_{1}}{\lambda_{1}D_{1}d_{2}} \Rightarrow \beta_{2} = 2.5 \times 10^{- 4}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.