The slits in Young?? double slit experiment are illuminated by light of wavelength $6000 ?$. If the path difference at the central bright fringe is zero, what is the path difference for light from the slits at the fourth bright fringe?
Updated On: Jul 7, 2022
$2.4\times 10^{-6}m$
$1.2\times 10^{-6}m$
$10^{-6}m$
$ 0.5 \times 10^{-6}m$
Hide Solution
Verified By Collegedunia
The Correct Option isA
Solution and Explanation
The path difference at the centre is zero.
The path difference at the $4^{th}$ bright fringe $= 4\lambda$
Here, $\lambda = 6000 \times 10^{-10}\, m$$\therefore$ Path difference $= 4 \times 6000 \times 10^{-10} $$= 24 \times 10^{-7}$$= 2.4 \times 10^{-6}\, m$
Considering two waves interfering at point P, having different distances. Consider a monochromatic light source ‘S’ kept at a relevant distance from two slits namely S1 and S2. S is at equal distance from S1 and S2. SO, we can assume that S1 and S2 are two coherent sources derived from S.
The light passes through these slits and falls on the screen that is kept at the distance D from both the slits S1 and S2. It is considered that d is the separation between both the slits. The S1 is opened, S2 is closed and the screen opposite to the S1 is closed, but the screen opposite to S2 is illuminating.
Thus, an interference pattern takes place when both the slits S1 and S2 are open. When the slit separation ‘d ‘and the screen distance D are kept unchanged, to reach point P the light waves from slits S1 and S2 must travel at different distances. It implies that there is a path difference in the Young double-slit experiment between the two slits S1 and S2.