Q. In Young’s double slit experiment when wavelength used is $6000\mathring{\mathrm{A}}$ and the screen is $40\ cm$ from the slits, the fringes are $0.012\ cm$ wide. What is the distance between the slits
A) $0.024\ cm$
B) $2.4\ cm$
C) $0.24\ cm$
D) $0.2\ cm$
Explanation
(d) $\beta = \frac{\lambda D}{d} \Rightarrow d = \frac{\lambda D}{\beta} = \frac{6000 \times 10^{- 10} \times (40 \times 10^{- 2})}{0.012 \times 10^{- 2}} = 0.2cm$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.