Question

The double slit experiment of Young has been shown in figure. Q is the position of the first bright fringe on the right side and P is the ${11}^{th}$ bright fringe on the other side as measured from Q. If wavelength of the light used is $6000A^{\circ },$ 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. $3.144\times {10}^{-7}m$

Answer 1

The correct option is A $6\times {10}^{-6}m$

From given data, we can observe that
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 waves reaching at P is
$=S_{1}B$ (10 fringes)
$=10\lambda =10\times 6000\times {10}^{-10}$
$=6\times {10}^{-6}m$