Question

A $YDSE$ apparatus is as shown in figure below. The condition for point $P$ to be a dark fringe is $(\lambda =$ wavelength of the light used)

• A. $(l_1-l_2)+(l_3-l_4)=n\lambda$
• B. $(l_1-l_3)+(l_3-l_4)=n\lambda$
• C. $(l_1+l_2)-(l_3+l_4)=\frac{(2n-1)\lambda }{2}$
• D. $(l_1+l_3)-(l_2+l_4)=\frac{(2n-1)\lambda }{2}$

Answer 1

The correct option is D $(l_1+l_3)-(l_2+l_4)=\frac{(2n-1)\lambda }{2}$

The path difference for the dark fringe is

$\Delta x=\frac{(2n-1)\lambda }{2}$

$\Rightarrow (S{S}_1+S_{1}P)-(S{S}_2+S_{2}P)=\frac{(2n-1)\lambda }{2}$

$\Rightarrow (l_1+l_3)-(l_2+l_4)=\frac{(2n-1)\lambda }{2}$

Hence, option $\left(\text{D}\right)$ is correct.