Question

The minimum value of d so that there is a dark fringe at O is $d_{min}$. The distance at which the next bright fringe is formed is x. Then

• A. $d_{min}=\sqrt{\lambda D}$
• B. $d_{min}=\sqrt{\frac{\lambda }{2}}$
• C. $x=\frac{d_{min}}{2}$
• D. $x=d_{min}$

Answer 1

Answer: (B), (D)

The correct options are
B $d_{min}=\sqrt{\frac{\lambda }{2}}$
D $x=d_{min}$

There is a dark fringe at O if the path difference $\Delta x=ABO-A{O}^{\prime }O=\frac{\lambda }{2}$
$\Delta x=2D{(1+\frac{d^2}{D^2})}^{\frac{1}{2}}-2D=2D(1+\frac{1}{2}\frac{d^2}{D^2})-2D=\frac{d^2}{D}\therefore \frac{\lambda }{2}=\frac{d_{min}^2}{D}Or,d_{min}=\sqrt{\frac{\lambda D}{2}}$
The bright fringe is formed at P if the path difference
= AO’P – ABP = 0
$\Rightarrow D+\sqrt{D^2+x^2}-D^2+d^2-\sqrt{D^2+(x-d{)}^2}=0\Rightarrow \frac{x^2}{2D}-\frac{d^2}{2D}-\frac{(x^2+d^2-2xd)}{2D}=0Givend=d_{min}$
On solving, $x=d_{min}=\sqrt{\frac{\lambda }{2}}$