Question

Two ideal slits $S_1$ and $S_2$ are at a distance $d$ apart and illuminated by light of wavelength $\lambda$ passing through an ideal source slit $S$ placed on the line through $S_2$ as shown. The distance between the planes of slits and the source slit is $D$ . A screen is held at a distance $D$ from the plane of the slots. The minimum value of $d$ for which there is darkness at $O$ is

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

Answer 1

The correct option is C $\sqrt{\frac{\lambda D}{2}}$

Path difference between the waves reaching at $P,\Delta ={\Delta }_1+{\Delta }_2$

where ${\Delta }_1=$ initial path difference ${\Delta }_2=$ Path difference between the waves after emerging from slits.

${\Delta }_1=S{S}_1-S{S}_2=\sqrt{D^2+d^2}-D$

and ${\Delta }_2=S_{1}O-S_{2}O=\sqrt{D^2+d^2}-D$

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

$=\frac{d^2}{D}$ (From Binomial expansion)

For obtaining dark at $O,\Delta$ must be equals to $(2n-1)\frac{\lambda }{2}$ i.e.

$\frac{d^2}{D}=(2n-1)\frac{\lambda }{2}\Rightarrow d\sqrt{\frac{(2n-1)\lambda D}{2}}$

For minimum distance $n-1$ so $d=\sqrt{\frac{\lambda D}{2}}$