Q. In a certain double slit experimental arrangement interference fringes of width $1.0\ mm$ each are observed when light of wavelength $5000\ \mathring{\mathrm{A}}$ is used. Keeping the set up unaltered, if the source is replaced by another source of wavelength $6000\ \mathring{\mathrm{A}}$, the fringe width will be
A) $0.5\ mm$
B) $1.0\ mm$
C) $1.2\ mm$
D) $1.5\ mm$
Explanation
(c) $\frac{\beta_{1}}{\beta_{2}} = \frac{\lambda_{1}}{\lambda_{2}}\ or\ \frac{1.0}{\beta_{2}} = \frac{5000}{6000}\ or\ \beta_{2} = \frac{6000}{5000} = 1.2\ mm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.