Question

Monochromatic light of wavelength $\lambda$ emerging from slit $S$ illuminates slits $S_1$ and $S_2$ which are placed with respect to $S$ as shown in figure. The distance $x$ and $D$ are large compared to the separation $d$ between the slits. If $x=D/2$, the minimum value of $d$ so that there is a dark fringe at the centre $P$ of the screen is

• A. $\sqrt{\lambda D}$
• B. $2\sqrt{\frac{\lambda D}{3}}$
  
• C. $\sqrt{\frac{2\lambda D}{3}}$
  
• D. $\sqrt{\frac{\lambda D}{3}}$
  

Answer 1

The correct option is D $\sqrt{\frac{\lambda D}{3}}$



The path difference on reading the point $P$

$\Delta x=(S{S}_2+S_{2}P)-(S{S}_1+S_{1}P)....\left(1\right)$

Here,

$S{S}_2=\sqrt{x^2+d^2}={\left[x^2(1+\frac{d^2}{x^2})\right]}^{1/2}=x(1+\frac{d^2}{2x^2})...\text{Using binomial expansion}$

$S_{2}P=\sqrt{D^2+d^2}={\left[D^2[1+\frac{d^2}{D^2}]\right]}^{1/2}=D[1+\frac{d^2}{2D^2}]...\text{Using binomial expansion}$

Equation $\left(1\right)$ becomes,

$\Delta x=x(1+\frac{d^2}{2x^2})+D(1+\frac{d^2}{2D^2})-(x+D)$

$=\frac{d^2}{2x}+\frac{d^2}{2D}=\frac{d^2}{2}[\frac{1}{x}+\frac{1}{D}]$

For dark fringe, $\Delta x=\frac{\lambda }{2}$, so the above equation becomes,

$\frac{\lambda }{2}=\frac{d^2}{2}[\frac{1}{x}+\frac{1}{D}]$

$\Rightarrow \lambda =d^2[\frac{2}{D}+\frac{1}{D}][\because x=D/2]$

$\Rightarrow d^2=\frac{\lambda D}{3}$

$d=\sqrt{\frac{\lambda D}{3}}$

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