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}} $



जेईई के लिए मॉक टेस्ट

एनसीईआरटी अध्याय वीडियो समाधान

दोहरा फलक