Q. In the figure is shown Young’s double slit experiment. $Q$ is the position of the first bright fringe on the right side of $O.P$ is the $11^{th}$ fringe on the other side, as measured from $Q$. If the wavelength of the light used is $6000 \times 10^{- 10}m$, then $S_{1}B$ will be equal to
A) $6 \times 10^{- 6}m$
B) $6.6 \times 10^{- 6}m$
C) $3.138 \times 10^{- 7}m$
D) $3.144 \times 10^{- 7}m$
Explanation
(a) $P$ is the position of 11$^{th}$ bright fringe from $Q$. From central position $O,P$ will be the position of 10$^{th}$ bright fringe
Path difference between the waves reaching at $P = S_{1}B = 10\lambda = 10 \times 6000 \times 10^{- 10} = 6 \times 10^{- 6}m$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.