Optics Question 158
Question: Two ideal slits S1 and S2 are at a distance d apart, and illuminated by light of wavelength l passing through an ideal source slit S placed on the line through S2 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 slits. The minimum value of d for which there is darkness at O is
Options:
A) $ \sqrt{\frac{3\lambda D}{2}} $
B) $ \sqrt{\lambda D} $
C) $ \sqrt{\frac{\lambda D}{2}} $
D) $ \sqrt{3\lambda D} $
Show Answer
Answer:
Correct Answer: C
Solution:
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}}=SS _{1}-SS _{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{ {{(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}} $
