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Two ideal slits \mathrm{S_1 } and \mathrm{S_2} are at a distance d apart and illuminated by light of wavelength \mathrm{\lambda} passing through an ideal source slit S placed on the line through \mathrm{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 slits. The minimum value of d for which there is darkness at O is:

Option: 1

\mathrm{\sqrt{\frac{3 \lambda \mathrm{D}}{2}}}


Option: 2

\mathrm{\sqrt{\lambda D}}


Option: 3

\mathrm{\sqrt{\frac{\lambda D}{2}}}


Option: 4

\mathrm{\sqrt{3 \lambda D}}


Answers (1)

best_answer

Path difference between the waves reaching at \mathrm{P, \Delta=\Delta_1+\Delta_2}

where, \mathrm{\Delta_1=} initial path difference

\mathrm{\Delta_2=} path difference between the waves after emerging from slits.

Now, \mathrm{\Delta_1=\mathrm{SS}_1-\mathrm{SS}_2=\sqrt{\left(\mathrm{D}^2+\mathrm{d}^2\right)}-\mathrm{D}}

and \mathrm{\Delta_2=\mathrm{S}_1 \mathrm{O}-\mathrm{S}_2 \mathrm{O}=\sqrt{\left(\mathrm{D}^2+\mathrm{d}^2\right)}-\mathrm{D}}

\mathrm{ \begin{aligned} & \therefore \Delta=2\left\{\sqrt{\left(D^2+d^2\right)}-D\right\}=2\left\{\left(D^2+d^2\right)^{1 / 2}-D\right\} \\ & =2\left\{\left(d+\frac{d^2}{2 D}\right)-D\right\} \text { from binomial expansion } \\ & =\frac{d^2}{D} \end{aligned} }

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

i.e.,\mathrm{ \frac{d^2}{D}=(2 n-1) \frac{\lambda}{2}}

\mathrm{ \therefore \quad \mathrm{d}^2=\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2} \quad \text { or } \quad \mathrm{d}=\sqrt{\frac{(2 \mathrm{n}-1) \lambda \mathrm{D}}{2}} }
For minimum distance \mathrm{\mathrm{n}=1}

So, \mathrm{\quad d=\sqrt{\left(\frac{\lambda D}{2}\right)}}

Posted by

Kuldeep Maurya

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