Question

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 11th 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. $3.144\times {10}^{-7}m$

Answer 1

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

P is the position of 11th bright fringe from Q. From central position O, P will be the position of 10th bright fringe.
Path difference between the waves reaching at P
$=S_{1}B=10\lambda =10\times 6000\times {10}^{-10}=6\times {10}^{-6}m$.